The length of the spring.

Question

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Answer 1

Question

Chapter 14, Problem 45PQ

(a)

To determine

The length of the spring.

(a)

Expert Solution

Answer to Problem 45PQ

The length of the spring is $8.62 m$ .

Explanation of Solution

Following figure gives rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 1

Following figure is the free body diagram of rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 2

Write the expression for the horizontal distance between one end to other end of the spring.

$xs=xp+xr$ (I)

Here, $xs$ is the horizontal distance between one end to other end of the spring, $xp$ is the distance between the pin in the spring and the pin in the rod and $xr$ is the horizontal distance between pin in the rod and end of the rod.

Write the expression for $xr$ .

$xr=Lrsin(41.6°)$ (II)

Here, $Lr$ is the length of the rod.

Write the expression for the vertical distance of the end of the spring from ground.

$yr=Lrcos(41.6°)$ (III)

Here, $yr$ is the vertical distance of the end of the rod from ground.

Write the expression for the length of the spring using Pythagoras theorem.

$L=xs2+ys2$ (IV)

Here, $L$ is the length of the spring and $ys$ is the vertical distance between the end of the spring from ground.

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (II) to get $xr$ .

$xr=(2.50 m)sin(41.6°)=1.66 m$

Substitute $2.50 m$ for $Lr$ in equation (III) to get $yr$ .

$yr=(2.50 m)cos(41.6°)=1.87 m$

This distance is also equal to vertical distance of the end of the spring from ground.

$ys=1.87 m$ .

Substitute $6.75 m$ for $xp$ and $1.66 m$ for $xr$ in equation (I) to get $xs$ .

$xs=6.75 m+1.66 m=8.41 m$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (IV) to get $L$ .

$L=(8.41 m)2+(1.87 m)2=8.62 m$

Therefore, the length of the spring is $8.62 m$ .

(b)

To determine

The weight of the bar.

(b)

Expert Solution

Answer to Problem 45PQ

The weight of the bar is $1.82 ×103 N$ .

Explanation of Solution

At equilibrium, the net torque acting on the bar around the bottom pivot must be zero.

Write the expression for the torque about pivot in the rod due to gravity.

$τg=Fgr⊥$ (V)

Here, $τg$ is the torque about pivot in the rod due to gravity, $Fg$ is the weight of the bar and $r⊥$ is the perpendicular distance between pivot of the rod and point where weight acts.

The direction of torque is into the page.

Using figure2, write the expression for the perpendicular distance between pivot of the rod and point where weight acts.

$r⊥=(Lr2)sin41.6°$ (VI)

Write the expression for the radial vector from the pivot point to the end of the bar where the spring acts.

$r→s=Lrsin41.6°i^+Lrcos41.6°j^$ (VII)

Here, $r→s$ is the radial vector from the pivot point to the end of the bar where the spring acts.

Write the expression for the relaxed length.

$Lrelaxed=xp2+Lr2$ (VIII)

Here, $Lrelaxed$ is the relaxed length.

The magnitude of spring force depends on the extension relative to the relaxed spring length.

Write the expression for the magnitude of spring force.

$Fs=kΔr$ (IX)

Here, $Fs$ is the spring force, $k$ is the spring constant and $Δr$ is the extension of spring relative to the relaxed spring length.

Write the expression for the extension of spring relative to the relaxed spring length.

$Δr=L−Lrelaxed$ (X)

From figure2, write the expression for the angle spring force makes below the horizontal.

$ϕ=tan−(ysxs)$ (XI)

Here, $ϕ$ is the angle that spring force makes below the horizontal.

Write the expression for the spring force as a vector.

$F→s=Fscosϕ(−i^)+Fssinϕ(−j^)$ (XII)

Here, $F→s$ is the vector form of spring force.

Write the expression for the torque on the rod due to spring force.

$τ→s=r→s×F→s$ (XIII)

Here, $τ→s$ is the torque on the rod due to spring force.

The direction of above torque is out of the page.

At equilibrium torque due to spring force and weight will cancel each other. Since, the directions of torques are opposite, their magnitude should be equal.

Write the equilibrium condition of the torques.

$τg=τs$ (XIV)

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (VI) to get $r⊥$ .

$r⊥=(2.50 m2)sin41.6°=(1.25 m)sin41.6°$

Substitute $(1.25 m)sin41.6°$ for $r⊥$ in equation (V) to get $τg$ .

$τg=Fg(2.50 m2)sin41.6°=Fg(1.25 m)sin41.6°$

Substitute $2.5 m$ for $Lr$ in equation (VII) to get $r→s$ .

$r→s=(2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^$ (XV)

Substitute $6.75 m$ for $xp$ and $2.50 m$ for $Lr$ in equation (VIII) to get $Lrelaxed$ .

$Lrelaxed=(6.75 m)2+(2.50 m)2=7.198 m$

Substitute $7.198 m$ for $Lrelaxed$ and $8.62 m$ for $L$ in equation (X) to get $Δr$ .

$Δr=8.62 m−7.198 m=1.42 m$

Substitute $1.42 m$ for $Δr$ and $750 N/m$ for $k$ in equation (IX) to get $Fs$ .

$Fs=(750 N/m)(1.42 m)=1.06×103 N$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (XI) to get $ϕ$ .

$ϕ=tan−(1.87 m8.41 m)=12.5 °$

Substitute $1.06×103 N$ for $Fs$ and $12.5 °$ for $ϕ$ in equation (XII)) to get $F→s$ .

$F→s=(1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^)$ (XVI)

Use equations (XV) and (XVI) in (XIII) to get $τ→s$ .

$τ→s=((2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^)×((1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^))=(−(2.5 m)sin41.6°×(1.06×103 N)sin(12.5 °)(i^×j^))+(−(2.5 m)cos41.6°×(1.06×103 N)cos(12.5 °)(j^×i^))=(1.51×103 N⋅m)k^$

Substitute $(1.51×103 N⋅m)$ for $τs$ and $Fg(1.25 m)sin41.6°$ for $τg$ in (XIV) to get $Fg$ .

$Fg(1.25 m)sin41.6°=(1.51×103 N⋅m)Fg=(1.51×103 N⋅m)(1.25 m)sin41.6°=1.82×103 N$

Therefore, The weight of the bar is $1.82 ×103 N$ .

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Answer 2

Question

Chapter 14, Problem 45PQ

(a)

To determine

The length of the spring.

(a)

Expert Solution

Answer to Problem 45PQ

The length of the spring is $8.62 m$ .

Explanation of Solution

Following figure gives rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 1

Following figure is the free body diagram of rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 2

Write the expression for the horizontal distance between one end to other end of the spring.

$xs=xp+xr$ (I)

Here, $xs$ is the horizontal distance between one end to other end of the spring, $xp$ is the distance between the pin in the spring and the pin in the rod and $xr$ is the horizontal distance between pin in the rod and end of the rod.

Write the expression for $xr$ .

$xr=Lrsin(41.6°)$ (II)

Here, $Lr$ is the length of the rod.

Write the expression for the vertical distance of the end of the spring from ground.

$yr=Lrcos(41.6°)$ (III)

Here, $yr$ is the vertical distance of the end of the rod from ground.

Write the expression for the length of the spring using Pythagoras theorem.

$L=xs2+ys2$ (IV)

Here, $L$ is the length of the spring and $ys$ is the vertical distance between the end of the spring from ground.

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (II) to get $xr$ .

$xr=(2.50 m)sin(41.6°)=1.66 m$

Substitute $2.50 m$ for $Lr$ in equation (III) to get $yr$ .

$yr=(2.50 m)cos(41.6°)=1.87 m$

This distance is also equal to vertical distance of the end of the spring from ground.

$ys=1.87 m$ .

Substitute $6.75 m$ for $xp$ and $1.66 m$ for $xr$ in equation (I) to get $xs$ .

$xs=6.75 m+1.66 m=8.41 m$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (IV) to get $L$ .

$L=(8.41 m)2+(1.87 m)2=8.62 m$

Therefore, the length of the spring is $8.62 m$ .

(b)

To determine

The weight of the bar.

(b)

Expert Solution

Answer to Problem 45PQ

The weight of the bar is $1.82 ×103 N$ .

Explanation of Solution

At equilibrium, the net torque acting on the bar around the bottom pivot must be zero.

Write the expression for the torque about pivot in the rod due to gravity.

$τg=Fgr⊥$ (V)

Here, $τg$ is the torque about pivot in the rod due to gravity, $Fg$ is the weight of the bar and $r⊥$ is the perpendicular distance between pivot of the rod and point where weight acts.

The direction of torque is into the page.

Using figure2, write the expression for the perpendicular distance between pivot of the rod and point where weight acts.

$r⊥=(Lr2)sin41.6°$ (VI)

Write the expression for the radial vector from the pivot point to the end of the bar where the spring acts.

$r→s=Lrsin41.6°i^+Lrcos41.6°j^$ (VII)

Here, $r→s$ is the radial vector from the pivot point to the end of the bar where the spring acts.

Write the expression for the relaxed length.

$Lrelaxed=xp2+Lr2$ (VIII)

Here, $Lrelaxed$ is the relaxed length.

The magnitude of spring force depends on the extension relative to the relaxed spring length.

Write the expression for the magnitude of spring force.

$Fs=kΔr$ (IX)

Here, $Fs$ is the spring force, $k$ is the spring constant and $Δr$ is the extension of spring relative to the relaxed spring length.

Write the expression for the extension of spring relative to the relaxed spring length.

$Δr=L−Lrelaxed$ (X)

From figure2, write the expression for the angle spring force makes below the horizontal.

$ϕ=tan−(ysxs)$ (XI)

Here, $ϕ$ is the angle that spring force makes below the horizontal.

Write the expression for the spring force as a vector.

$F→s=Fscosϕ(−i^)+Fssinϕ(−j^)$ (XII)

Here, $F→s$ is the vector form of spring force.

Write the expression for the torque on the rod due to spring force.

$τ→s=r→s×F→s$ (XIII)

Here, $τ→s$ is the torque on the rod due to spring force.

The direction of above torque is out of the page.

At equilibrium torque due to spring force and weight will cancel each other. Since, the directions of torques are opposite, their magnitude should be equal.

Write the equilibrium condition of the torques.

$τg=τs$ (XIV)

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (VI) to get $r⊥$ .

$r⊥=(2.50 m2)sin41.6°=(1.25 m)sin41.6°$

Substitute $(1.25 m)sin41.6°$ for $r⊥$ in equation (V) to get $τg$ .

$τg=Fg(2.50 m2)sin41.6°=Fg(1.25 m)sin41.6°$

Substitute $2.5 m$ for $Lr$ in equation (VII) to get $r→s$ .

$r→s=(2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^$ (XV)

Substitute $6.75 m$ for $xp$ and $2.50 m$ for $Lr$ in equation (VIII) to get $Lrelaxed$ .

$Lrelaxed=(6.75 m)2+(2.50 m)2=7.198 m$

Substitute $7.198 m$ for $Lrelaxed$ and $8.62 m$ for $L$ in equation (X) to get $Δr$ .

$Δr=8.62 m−7.198 m=1.42 m$

Substitute $1.42 m$ for $Δr$ and $750 N/m$ for $k$ in equation (IX) to get $Fs$ .

$Fs=(750 N/m)(1.42 m)=1.06×103 N$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (XI) to get $ϕ$ .

$ϕ=tan−(1.87 m8.41 m)=12.5 °$

Substitute $1.06×103 N$ for $Fs$ and $12.5 °$ for $ϕ$ in equation (XII)) to get $F→s$ .

$F→s=(1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^)$ (XVI)

Use equations (XV) and (XVI) in (XIII) to get $τ→s$ .

$τ→s=((2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^)×((1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^))=(−(2.5 m)sin41.6°×(1.06×103 N)sin(12.5 °)(i^×j^))+(−(2.5 m)cos41.6°×(1.06×103 N)cos(12.5 °)(j^×i^))=(1.51×103 N⋅m)k^$

Substitute $(1.51×103 N⋅m)$ for $τs$ and $Fg(1.25 m)sin41.6°$ for $τg$ in (XIV) to get $Fg$ .

$Fg(1.25 m)sin41.6°=(1.51×103 N⋅m)Fg=(1.51×103 N⋅m)(1.25 m)sin41.6°=1.82×103 N$

Therefore, The weight of the bar is $1.82 ×103 N$ .

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Answer 3

Question

Chapter 14, Problem 45PQ

(a)

To determine

The length of the spring.

(a)

Expert Solution

Answer to Problem 45PQ

The length of the spring is $8.62 m$ .

Explanation of Solution

Following figure gives rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 1

Following figure is the free body diagram of rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 2

Write the expression for the horizontal distance between one end to other end of the spring.

$xs=xp+xr$ (I)

Here, $xs$ is the horizontal distance between one end to other end of the spring, $xp$ is the distance between the pin in the spring and the pin in the rod and $xr$ is the horizontal distance between pin in the rod and end of the rod.

Write the expression for $xr$ .

$xr=Lrsin(41.6°)$ (II)

Here, $Lr$ is the length of the rod.

Write the expression for the vertical distance of the end of the spring from ground.

$yr=Lrcos(41.6°)$ (III)

Here, $yr$ is the vertical distance of the end of the rod from ground.

Write the expression for the length of the spring using Pythagoras theorem.

$L=xs2+ys2$ (IV)

Here, $L$ is the length of the spring and $ys$ is the vertical distance between the end of the spring from ground.

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (II) to get $xr$ .

$xr=(2.50 m)sin(41.6°)=1.66 m$

Substitute $2.50 m$ for $Lr$ in equation (III) to get $yr$ .

$yr=(2.50 m)cos(41.6°)=1.87 m$

This distance is also equal to vertical distance of the end of the spring from ground.

$ys=1.87 m$ .

Substitute $6.75 m$ for $xp$ and $1.66 m$ for $xr$ in equation (I) to get $xs$ .

$xs=6.75 m+1.66 m=8.41 m$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (IV) to get $L$ .

$L=(8.41 m)2+(1.87 m)2=8.62 m$

Therefore, the length of the spring is $8.62 m$ .

(b)

To determine

The weight of the bar.

(b)

Expert Solution

Answer to Problem 45PQ

The weight of the bar is $1.82 ×103 N$ .

Explanation of Solution

At equilibrium, the net torque acting on the bar around the bottom pivot must be zero.

Write the expression for the torque about pivot in the rod due to gravity.

$τg=Fgr⊥$ (V)

Here, $τg$ is the torque about pivot in the rod due to gravity, $Fg$ is the weight of the bar and $r⊥$ is the perpendicular distance between pivot of the rod and point where weight acts.

The direction of torque is into the page.

Using figure2, write the expression for the perpendicular distance between pivot of the rod and point where weight acts.

$r⊥=(Lr2)sin41.6°$ (VI)

Write the expression for the radial vector from the pivot point to the end of the bar where the spring acts.

$r→s=Lrsin41.6°i^+Lrcos41.6°j^$ (VII)

Here, $r→s$ is the radial vector from the pivot point to the end of the bar where the spring acts.

Write the expression for the relaxed length.

$Lrelaxed=xp2+Lr2$ (VIII)

Here, $Lrelaxed$ is the relaxed length.

The magnitude of spring force depends on the extension relative to the relaxed spring length.

Write the expression for the magnitude of spring force.

$Fs=kΔr$ (IX)

Here, $Fs$ is the spring force, $k$ is the spring constant and $Δr$ is the extension of spring relative to the relaxed spring length.

Write the expression for the extension of spring relative to the relaxed spring length.

$Δr=L−Lrelaxed$ (X)

From figure2, write the expression for the angle spring force makes below the horizontal.

$ϕ=tan−(ysxs)$ (XI)

Here, $ϕ$ is the angle that spring force makes below the horizontal.

Write the expression for the spring force as a vector.

$F→s=Fscosϕ(−i^)+Fssinϕ(−j^)$ (XII)

Here, $F→s$ is the vector form of spring force.

Write the expression for the torque on the rod due to spring force.

$τ→s=r→s×F→s$ (XIII)

Here, $τ→s$ is the torque on the rod due to spring force.

The direction of above torque is out of the page.

At equilibrium torque due to spring force and weight will cancel each other. Since, the directions of torques are opposite, their magnitude should be equal.

Write the equilibrium condition of the torques.

$τg=τs$ (XIV)

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (VI) to get $r⊥$ .

$r⊥=(2.50 m2)sin41.6°=(1.25 m)sin41.6°$

Substitute $(1.25 m)sin41.6°$ for $r⊥$ in equation (V) to get $τg$ .

$τg=Fg(2.50 m2)sin41.6°=Fg(1.25 m)sin41.6°$

Substitute $2.5 m$ for $Lr$ in equation (VII) to get $r→s$ .

$r→s=(2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^$ (XV)

Substitute $6.75 m$ for $xp$ and $2.50 m$ for $Lr$ in equation (VIII) to get $Lrelaxed$ .

$Lrelaxed=(6.75 m)2+(2.50 m)2=7.198 m$

Substitute $7.198 m$ for $Lrelaxed$ and $8.62 m$ for $L$ in equation (X) to get $Δr$ .

$Δr=8.62 m−7.198 m=1.42 m$

Substitute $1.42 m$ for $Δr$ and $750 N/m$ for $k$ in equation (IX) to get $Fs$ .

$Fs=(750 N/m)(1.42 m)=1.06×103 N$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (XI) to get $ϕ$ .

$ϕ=tan−(1.87 m8.41 m)=12.5 °$

Substitute $1.06×103 N$ for $Fs$ and $12.5 °$ for $ϕ$ in equation (XII)) to get $F→s$ .

$F→s=(1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^)$ (XVI)

Use equations (XV) and (XVI) in (XIII) to get $τ→s$ .

$τ→s=((2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^)×((1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^))=(−(2.5 m)sin41.6°×(1.06×103 N)sin(12.5 °)(i^×j^))+(−(2.5 m)cos41.6°×(1.06×103 N)cos(12.5 °)(j^×i^))=(1.51×103 N⋅m)k^$

Substitute $(1.51×103 N⋅m)$ for $τs$ and $Fg(1.25 m)sin41.6°$ for $τg$ in (XIV) to get $Fg$ .

$Fg(1.25 m)sin41.6°=(1.51×103 N⋅m)Fg=(1.51×103 N⋅m)(1.25 m)sin41.6°=1.82×103 N$

Therefore, The weight of the bar is $1.82 ×103 N$ .

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Answer 4

Question

Chapter 14, Problem 45PQ

(a)

To determine

The length of the spring.

(a)

Expert Solution

Answer to Problem 45PQ

The length of the spring is $8.62 m$ .

Explanation of Solution

Following figure gives rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 1

Following figure is the free body diagram of rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 2

Write the expression for the horizontal distance between one end to other end of the spring.

$xs=xp+xr$ (I)

Here, $xs$ is the horizontal distance between one end to other end of the spring, $xp$ is the distance between the pin in the spring and the pin in the rod and $xr$ is the horizontal distance between pin in the rod and end of the rod.

Write the expression for $xr$ .

$xr=Lrsin(41.6°)$ (II)

Here, $Lr$ is the length of the rod.

Write the expression for the vertical distance of the end of the spring from ground.

$yr=Lrcos(41.6°)$ (III)

Here, $yr$ is the vertical distance of the end of the rod from ground.

Write the expression for the length of the spring using Pythagoras theorem.

$L=xs2+ys2$ (IV)

Here, $L$ is the length of the spring and $ys$ is the vertical distance between the end of the spring from ground.

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (II) to get $xr$ .

$xr=(2.50 m)sin(41.6°)=1.66 m$

Substitute $2.50 m$ for $Lr$ in equation (III) to get $yr$ .

$yr=(2.50 m)cos(41.6°)=1.87 m$

This distance is also equal to vertical distance of the end of the spring from ground.

$ys=1.87 m$ .

Substitute $6.75 m$ for $xp$ and $1.66 m$ for $xr$ in equation (I) to get $xs$ .

$xs=6.75 m+1.66 m=8.41 m$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (IV) to get $L$ .

$L=(8.41 m)2+(1.87 m)2=8.62 m$

Therefore, the length of the spring is $8.62 m$ .

(b)

To determine

The weight of the bar.

(b)

Expert Solution

Answer to Problem 45PQ

The weight of the bar is $1.82 ×103 N$ .

Explanation of Solution

At equilibrium, the net torque acting on the bar around the bottom pivot must be zero.

Write the expression for the torque about pivot in the rod due to gravity.

$τg=Fgr⊥$ (V)

Here, $τg$ is the torque about pivot in the rod due to gravity, $Fg$ is the weight of the bar and $r⊥$ is the perpendicular distance between pivot of the rod and point where weight acts.

The direction of torque is into the page.

Using figure2, write the expression for the perpendicular distance between pivot of the rod and point where weight acts.

$r⊥=(Lr2)sin41.6°$ (VI)

Write the expression for the radial vector from the pivot point to the end of the bar where the spring acts.

$r→s=Lrsin41.6°i^+Lrcos41.6°j^$ (VII)

Here, $r→s$ is the radial vector from the pivot point to the end of the bar where the spring acts.

Write the expression for the relaxed length.

$Lrelaxed=xp2+Lr2$ (VIII)

Here, $Lrelaxed$ is the relaxed length.

The magnitude of spring force depends on the extension relative to the relaxed spring length.

Write the expression for the magnitude of spring force.

$Fs=kΔr$ (IX)

Here, $Fs$ is the spring force, $k$ is the spring constant and $Δr$ is the extension of spring relative to the relaxed spring length.

Write the expression for the extension of spring relative to the relaxed spring length.

$Δr=L−Lrelaxed$ (X)

From figure2, write the expression for the angle spring force makes below the horizontal.

$ϕ=tan−(ysxs)$ (XI)

Here, $ϕ$ is the angle that spring force makes below the horizontal.

Write the expression for the spring force as a vector.

$F→s=Fscosϕ(−i^)+Fssinϕ(−j^)$ (XII)

Here, $F→s$ is the vector form of spring force.

Write the expression for the torque on the rod due to spring force.

$τ→s=r→s×F→s$ (XIII)

Here, $τ→s$ is the torque on the rod due to spring force.

The direction of above torque is out of the page.

At equilibrium torque due to spring force and weight will cancel each other. Since, the directions of torques are opposite, their magnitude should be equal.

Write the equilibrium condition of the torques.

$τg=τs$ (XIV)

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (VI) to get $r⊥$ .

$r⊥=(2.50 m2)sin41.6°=(1.25 m)sin41.6°$

Substitute $(1.25 m)sin41.6°$ for $r⊥$ in equation (V) to get $τg$ .

$τg=Fg(2.50 m2)sin41.6°=Fg(1.25 m)sin41.6°$

Substitute $2.5 m$ for $Lr$ in equation (VII) to get $r→s$ .

$r→s=(2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^$ (XV)

Substitute $6.75 m$ for $xp$ and $2.50 m$ for $Lr$ in equation (VIII) to get $Lrelaxed$ .

$Lrelaxed=(6.75 m)2+(2.50 m)2=7.198 m$

Substitute $7.198 m$ for $Lrelaxed$ and $8.62 m$ for $L$ in equation (X) to get $Δr$ .

$Δr=8.62 m−7.198 m=1.42 m$

Substitute $1.42 m$ for $Δr$ and $750 N/m$ for $k$ in equation (IX) to get $Fs$ .

$Fs=(750 N/m)(1.42 m)=1.06×103 N$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (XI) to get $ϕ$ .

$ϕ=tan−(1.87 m8.41 m)=12.5 °$

Substitute $1.06×103 N$ for $Fs$ and $12.5 °$ for $ϕ$ in equation (XII)) to get $F→s$ .

$F→s=(1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^)$ (XVI)

Use equations (XV) and (XVI) in (XIII) to get $τ→s$ .

$τ→s=((2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^)×((1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^))=(−(2.5 m)sin41.6°×(1.06×103 N)sin(12.5 °)(i^×j^))+(−(2.5 m)cos41.6°×(1.06×103 N)cos(12.5 °)(j^×i^))=(1.51×103 N⋅m)k^$

Substitute $(1.51×103 N⋅m)$ for $τs$ and $Fg(1.25 m)sin41.6°$ for $τg$ in (XIV) to get $Fg$ .

$Fg(1.25 m)sin41.6°=(1.51×103 N⋅m)Fg=(1.51×103 N⋅m)(1.25 m)sin41.6°=1.82×103 N$

Therefore, The weight of the bar is $1.82 ×103 N$ .

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Answer 5

Chapter 14, Problem 45PQ

(a)

To determine

The length of the spring.

(a)

Expert Solution

Answer to Problem 45PQ

The length of the spring is $8.62 m$ .

Explanation of Solution

Following figure gives rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 1

Following figure is the free body diagram of rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 2

Write the expression for the horizontal distance between one end to other end of the spring.

$xs=xp+xr$ (I)

Here, $xs$ is the horizontal distance between one end to other end of the spring, $xp$ is the distance between the pin in the spring and the pin in the rod and $xr$ is the horizontal distance between pin in the rod and end of the rod.

Write the expression for $xr$ .

$xr=Lrsin(41.6°)$ (II)

Here, $Lr$ is the length of the rod.

Write the expression for the vertical distance of the end of the spring from ground.

$yr=Lrcos(41.6°)$ (III)

Here, $yr$ is the vertical distance of the end of the rod from ground.

Write the expression for the length of the spring using Pythagoras theorem.

$L=xs2+ys2$ (IV)

Here, $L$ is the length of the spring and $ys$ is the vertical distance between the end of the spring from ground.

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (II) to get $xr$ .

$xr=(2.50 m)sin(41.6°)=1.66 m$

Substitute $2.50 m$ for $Lr$ in equation (III) to get $yr$ .

$yr=(2.50 m)cos(41.6°)=1.87 m$

This distance is also equal to vertical distance of the end of the spring from ground.

$ys=1.87 m$ .

Substitute $6.75 m$ for $xp$ and $1.66 m$ for $xr$ in equation (I) to get $xs$ .

$xs=6.75 m+1.66 m=8.41 m$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (IV) to get $L$ .

$L=(8.41 m)2+(1.87 m)2=8.62 m$

Therefore, the length of the spring is $8.62 m$ .

(b)

To determine

The weight of the bar.

(b)

Expert Solution

Answer to Problem 45PQ

The weight of the bar is $1.82 ×103 N$ .

Explanation of Solution

At equilibrium, the net torque acting on the bar around the bottom pivot must be zero.

Write the expression for the torque about pivot in the rod due to gravity.

$τg=Fgr⊥$ (V)

Here, $τg$ is the torque about pivot in the rod due to gravity, $Fg$ is the weight of the bar and $r⊥$ is the perpendicular distance between pivot of the rod and point where weight acts.

The direction of torque is into the page.

Using figure2, write the expression for the perpendicular distance between pivot of the rod and point where weight acts.

$r⊥=(Lr2)sin41.6°$ (VI)

Write the expression for the radial vector from the pivot point to the end of the bar where the spring acts.

$r→s=Lrsin41.6°i^+Lrcos41.6°j^$ (VII)

Here, $r→s$ is the radial vector from the pivot point to the end of the bar where the spring acts.

Write the expression for the relaxed length.

$Lrelaxed=xp2+Lr2$ (VIII)

Here, $Lrelaxed$ is the relaxed length.

The magnitude of spring force depends on the extension relative to the relaxed spring length.

Write the expression for the magnitude of spring force.

$Fs=kΔr$ (IX)

Here, $Fs$ is the spring force, $k$ is the spring constant and $Δr$ is the extension of spring relative to the relaxed spring length.

Write the expression for the extension of spring relative to the relaxed spring length.

$Δr=L−Lrelaxed$ (X)

From figure2, write the expression for the angle spring force makes below the horizontal.

$ϕ=tan−(ysxs)$ (XI)

Here, $ϕ$ is the angle that spring force makes below the horizontal.

Write the expression for the spring force as a vector.

$F→s=Fscosϕ(−i^)+Fssinϕ(−j^)$ (XII)

Here, $F→s$ is the vector form of spring force.

Write the expression for the torque on the rod due to spring force.

$τ→s=r→s×F→s$ (XIII)

Here, $τ→s$ is the torque on the rod due to spring force.

The direction of above torque is out of the page.

At equilibrium torque due to spring force and weight will cancel each other. Since, the directions of torques are opposite, their magnitude should be equal.

Write the equilibrium condition of the torques.

$τg=τs$ (XIV)

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (VI) to get $r⊥$ .

$r⊥=(2.50 m2)sin41.6°=(1.25 m)sin41.6°$

Substitute $(1.25 m)sin41.6°$ for $r⊥$ in equation (V) to get $τg$ .

$τg=Fg(2.50 m2)sin41.6°=Fg(1.25 m)sin41.6°$

Substitute $2.5 m$ for $Lr$ in equation (VII) to get $r→s$ .

$r→s=(2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^$ (XV)

Substitute $6.75 m$ for $xp$ and $2.50 m$ for $Lr$ in equation (VIII) to get $Lrelaxed$ .

$Lrelaxed=(6.75 m)2+(2.50 m)2=7.198 m$

Substitute $7.198 m$ for $Lrelaxed$ and $8.62 m$ for $L$ in equation (X) to get $Δr$ .

$Δr=8.62 m−7.198 m=1.42 m$

Substitute $1.42 m$ for $Δr$ and $750 N/m$ for $k$ in equation (IX) to get $Fs$ .

$Fs=(750 N/m)(1.42 m)=1.06×103 N$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (XI) to get $ϕ$ .

$ϕ=tan−(1.87 m8.41 m)=12.5 °$

Substitute $1.06×103 N$ for $Fs$ and $12.5 °$ for $ϕ$ in equation (XII)) to get $F→s$ .

$F→s=(1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^)$ (XVI)

Use equations (XV) and (XVI) in (XIII) to get $τ→s$ .

$τ→s=((2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^)×((1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^))=(−(2.5 m)sin41.6°×(1.06×103 N)sin(12.5 °)(i^×j^))+(−(2.5 m)cos41.6°×(1.06×103 N)cos(12.5 °)(j^×i^))=(1.51×103 N⋅m)k^$

Substitute $(1.51×103 N⋅m)$ for $τs$ and $Fg(1.25 m)sin41.6°$ for $τg$ in (XIV) to get $Fg$ .

$Fg(1.25 m)sin41.6°=(1.51×103 N⋅m)Fg=(1.51×103 N⋅m)(1.25 m)sin41.6°=1.82×103 N$

Therefore, The weight of the bar is $1.82 ×103 N$ .

Answer 6

Chapter 14, Problem 45PQ

(a)

To determine

The length of the spring.

(a)

Expert Solution

Answer to Problem 45PQ

The length of the spring is $8.62 m$ .

Explanation of Solution

Following figure gives rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 1

Following figure is the free body diagram of rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 2

Write the expression for the horizontal distance between one end to other end of the spring.

$xs=xp+xr$ (I)

Here, $xs$ is the horizontal distance between one end to other end of the spring, $xp$ is the distance between the pin in the spring and the pin in the rod and $xr$ is the horizontal distance between pin in the rod and end of the rod.

Write the expression for $xr$ .

$xr=Lrsin(41.6°)$ (II)

Here, $Lr$ is the length of the rod.

Write the expression for the vertical distance of the end of the spring from ground.

$yr=Lrcos(41.6°)$ (III)

Here, $yr$ is the vertical distance of the end of the rod from ground.

Write the expression for the length of the spring using Pythagoras theorem.

$L=xs2+ys2$ (IV)

Here, $L$ is the length of the spring and $ys$ is the vertical distance between the end of the spring from ground.

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (II) to get $xr$ .

$xr=(2.50 m)sin(41.6°)=1.66 m$

Substitute $2.50 m$ for $Lr$ in equation (III) to get $yr$ .

$yr=(2.50 m)cos(41.6°)=1.87 m$

This distance is also equal to vertical distance of the end of the spring from ground.

$ys=1.87 m$ .

Substitute $6.75 m$ for $xp$ and $1.66 m$ for $xr$ in equation (I) to get $xs$ .

$xs=6.75 m+1.66 m=8.41 m$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (IV) to get $L$ .

$L=(8.41 m)2+(1.87 m)2=8.62 m$

Therefore, the length of the spring is $8.62 m$ .

Answer 7

Chapter 14, Problem 45PQ

(a)

To determine

The length of the spring.

Answer 8

(a)

Expert Solution

Answer to Problem 45PQ

The length of the spring is $8.62 m$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Following figure gives rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 1

Following figure is the free body diagram of rod and spring system.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 14, Problem 45PQ , additional homework tip 2

Write the expression for the horizontal distance between one end to other end of the spring.

$xs=xp+xr$ (I)

Here, $xs$ is the horizontal distance between one end to other end of the spring, $xp$ is the distance between the pin in the spring and the pin in the rod and $xr$ is the horizontal distance between pin in the rod and end of the rod.

Write the expression for $xr$ .

$xr=Lrsin(41.6°)$ (II)

Here, $Lr$ is the length of the rod.

Write the expression for the vertical distance of the end of the spring from ground.

$yr=Lrcos(41.6°)$ (III)

Here, $yr$ is the vertical distance of the end of the rod from ground.

Write the expression for the length of the spring using Pythagoras theorem.

$L=xs2+ys2$ (IV)

Here, $L$ is the length of the spring and $ys$ is the vertical distance between the end of the spring from ground.

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (II) to get $xr$ .

$xr=(2.50 m)sin(41.6°)=1.66 m$

Substitute $2.50 m$ for $Lr$ in equation (III) to get $yr$ .

$yr=(2.50 m)cos(41.6°)=1.87 m$

This distance is also equal to vertical distance of the end of the spring from ground.

$ys=1.87 m$ .

Substitute $6.75 m$ for $xp$ and $1.66 m$ for $xr$ in equation (I) to get $xs$ .

$xs=6.75 m+1.66 m=8.41 m$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (IV) to get $L$ .

$L=(8.41 m)2+(1.87 m)2=8.62 m$

Therefore, the length of the spring is $8.62 m$ .

Answer 13

(b)

To determine

The weight of the bar.

(b)

Expert Solution

Answer to Problem 45PQ

The weight of the bar is $1.82 ×103 N$ .

Explanation of Solution

At equilibrium, the net torque acting on the bar around the bottom pivot must be zero.

Write the expression for the torque about pivot in the rod due to gravity.

$τg=Fgr⊥$ (V)

Here, $τg$ is the torque about pivot in the rod due to gravity, $Fg$ is the weight of the bar and $r⊥$ is the perpendicular distance between pivot of the rod and point where weight acts.

The direction of torque is into the page.

Using figure2, write the expression for the perpendicular distance between pivot of the rod and point where weight acts.

$r⊥=(Lr2)sin41.6°$ (VI)

Write the expression for the radial vector from the pivot point to the end of the bar where the spring acts.

$r→s=Lrsin41.6°i^+Lrcos41.6°j^$ (VII)

Here, $r→s$ is the radial vector from the pivot point to the end of the bar where the spring acts.

Write the expression for the relaxed length.

$Lrelaxed=xp2+Lr2$ (VIII)

Here, $Lrelaxed$ is the relaxed length.

The magnitude of spring force depends on the extension relative to the relaxed spring length.

Write the expression for the magnitude of spring force.

$Fs=kΔr$ (IX)

Here, $Fs$ is the spring force, $k$ is the spring constant and $Δr$ is the extension of spring relative to the relaxed spring length.

Write the expression for the extension of spring relative to the relaxed spring length.

$Δr=L−Lrelaxed$ (X)

From figure2, write the expression for the angle spring force makes below the horizontal.

$ϕ=tan−(ysxs)$ (XI)

Here, $ϕ$ is the angle that spring force makes below the horizontal.

Write the expression for the spring force as a vector.

$F→s=Fscosϕ(−i^)+Fssinϕ(−j^)$ (XII)

Here, $F→s$ is the vector form of spring force.

Write the expression for the torque on the rod due to spring force.

$τ→s=r→s×F→s$ (XIII)

Here, $τ→s$ is the torque on the rod due to spring force.

The direction of above torque is out of the page.

At equilibrium torque due to spring force and weight will cancel each other. Since, the directions of torques are opposite, their magnitude should be equal.

Write the equilibrium condition of the torques.

$τg=τs$ (XIV)

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (VI) to get $r⊥$ .

$r⊥=(2.50 m2)sin41.6°=(1.25 m)sin41.6°$

Substitute $(1.25 m)sin41.6°$ for $r⊥$ in equation (V) to get $τg$ .

$τg=Fg(2.50 m2)sin41.6°=Fg(1.25 m)sin41.6°$

Substitute $2.5 m$ for $Lr$ in equation (VII) to get $r→s$ .

$r→s=(2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^$ (XV)

Substitute $6.75 m$ for $xp$ and $2.50 m$ for $Lr$ in equation (VIII) to get $Lrelaxed$ .

$Lrelaxed=(6.75 m)2+(2.50 m)2=7.198 m$

Substitute $7.198 m$ for $Lrelaxed$ and $8.62 m$ for $L$ in equation (X) to get $Δr$ .

$Δr=8.62 m−7.198 m=1.42 m$

Substitute $1.42 m$ for $Δr$ and $750 N/m$ for $k$ in equation (IX) to get $Fs$ .

$Fs=(750 N/m)(1.42 m)=1.06×103 N$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (XI) to get $ϕ$ .

$ϕ=tan−(1.87 m8.41 m)=12.5 °$

Substitute $1.06×103 N$ for $Fs$ and $12.5 °$ for $ϕ$ in equation (XII)) to get $F→s$ .

$F→s=(1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^)$ (XVI)

Use equations (XV) and (XVI) in (XIII) to get $τ→s$ .

$τ→s=((2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^)×((1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^))=(−(2.5 m)sin41.6°×(1.06×103 N)sin(12.5 °)(i^×j^))+(−(2.5 m)cos41.6°×(1.06×103 N)cos(12.5 °)(j^×i^))=(1.51×103 N⋅m)k^$

Substitute $(1.51×103 N⋅m)$ for $τs$ and $Fg(1.25 m)sin41.6°$ for $τg$ in (XIV) to get $Fg$ .

$Fg(1.25 m)sin41.6°=(1.51×103 N⋅m)Fg=(1.51×103 N⋅m)(1.25 m)sin41.6°=1.82×103 N$

Therefore, The weight of the bar is $1.82 ×103 N$ .

Answer 14

(b)

To determine

The weight of the bar.

Answer 15

(b)

Expert Solution

Answer to Problem 45PQ

The weight of the bar is $1.82 ×103 N$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

At equilibrium, the net torque acting on the bar around the bottom pivot must be zero.

Write the expression for the torque about pivot in the rod due to gravity.

$τg=Fgr⊥$ (V)

Here, $τg$ is the torque about pivot in the rod due to gravity, $Fg$ is the weight of the bar and $r⊥$ is the perpendicular distance between pivot of the rod and point where weight acts.

The direction of torque is into the page.

Using figure2, write the expression for the perpendicular distance between pivot of the rod and point where weight acts.

$r⊥=(Lr2)sin41.6°$ (VI)

Write the expression for the radial vector from the pivot point to the end of the bar where the spring acts.

$r→s=Lrsin41.6°i^+Lrcos41.6°j^$ (VII)

Here, $r→s$ is the radial vector from the pivot point to the end of the bar where the spring acts.

Write the expression for the relaxed length.

$Lrelaxed=xp2+Lr2$ (VIII)

Here, $Lrelaxed$ is the relaxed length.

The magnitude of spring force depends on the extension relative to the relaxed spring length.

Write the expression for the magnitude of spring force.

$Fs=kΔr$ (IX)

Here, $Fs$ is the spring force, $k$ is the spring constant and $Δr$ is the extension of spring relative to the relaxed spring length.

Write the expression for the extension of spring relative to the relaxed spring length.

$Δr=L−Lrelaxed$ (X)

From figure2, write the expression for the angle spring force makes below the horizontal.

$ϕ=tan−(ysxs)$ (XI)

Here, $ϕ$ is the angle that spring force makes below the horizontal.

Write the expression for the spring force as a vector.

$F→s=Fscosϕ(−i^)+Fssinϕ(−j^)$ (XII)

Here, $F→s$ is the vector form of spring force.

Write the expression for the torque on the rod due to spring force.

$τ→s=r→s×F→s$ (XIII)

Here, $τ→s$ is the torque on the rod due to spring force.

The direction of above torque is out of the page.

At equilibrium torque due to spring force and weight will cancel each other. Since, the directions of torques are opposite, their magnitude should be equal.

Write the equilibrium condition of the torques.

$τg=τs$ (XIV)

Conclusion:

Substitute $2.50 m$ for $Lr$ in equation (VI) to get $r⊥$ .

$r⊥=(2.50 m2)sin41.6°=(1.25 m)sin41.6°$

Substitute $(1.25 m)sin41.6°$ for $r⊥$ in equation (V) to get $τg$ .

$τg=Fg(2.50 m2)sin41.6°=Fg(1.25 m)sin41.6°$

Substitute $2.5 m$ for $Lr$ in equation (VII) to get $r→s$ .

$r→s=(2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^$ (XV)

Substitute $6.75 m$ for $xp$ and $2.50 m$ for $Lr$ in equation (VIII) to get $Lrelaxed$ .

$Lrelaxed=(6.75 m)2+(2.50 m)2=7.198 m$

Substitute $7.198 m$ for $Lrelaxed$ and $8.62 m$ for $L$ in equation (X) to get $Δr$ .

$Δr=8.62 m−7.198 m=1.42 m$

Substitute $1.42 m$ for $Δr$ and $750 N/m$ for $k$ in equation (IX) to get $Fs$ .

$Fs=(750 N/m)(1.42 m)=1.06×103 N$

Substitute $8.41 m$ for $xs$ and $1.87 m$ for $ys$ in equation (XI) to get $ϕ$ .

$ϕ=tan−(1.87 m8.41 m)=12.5 °$

Substitute $1.06×103 N$ for $Fs$ and $12.5 °$ for $ϕ$ in equation (XII)) to get $F→s$ .

$F→s=(1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^)$ (XVI)

Use equations (XV) and (XVI) in (XIII) to get $τ→s$ .

$τ→s=((2.5 m)sin41.6°i^+(2.5 m)cos41.6°j^)×((1.06×103 N)scos(12.5 °)(−i^)+(1.06×103 N)sin(12.5 °)(−j^))=(−(2.5 m)sin41.6°×(1.06×103 N)sin(12.5 °)(i^×j^))+(−(2.5 m)cos41.6°×(1.06×103 N)cos(12.5 °)(j^×i^))=(1.51×103 N⋅m)k^$