EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
16th Edition
ISBN: 8220100546716
Author: Katz
Publisher: CENGAGE L
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Chapter 14, Problem 36PQ

A square plate with sides of length 4.0 m can rotate about an axle passing through its center of mass and perpendicular to the plate as shown in Figure P14.36. There are four forces acting on the plate at different points. The rotational inertia of the plate is 24 kg·m2. Is the plate in equilibrium?

Chapter 14, Problem 36PQ, A square plate with sides of length 4.0 m can rotate about an axle passing through its center of

FIGURE P14.36

Expert Solution & Answer
Check Mark
To determine

Whether the plate is in equilibrium, if rotational inertia of the plate is 24kgm2 .

Answer to Problem 36PQ

Since net torque acting on the plate is zero, the pate is in equilibrium.

Explanation of Solution

Take right direction as +x direction, upward direction as +y direction with center of mass of the plate as origin.

From figure P14.36, write the expression for the vector form of force of magnitude 60.0N .

  F1=(60.0N)j^                                                                                                        (I)

Here, F1 is the force of magnitude 60.0N and j^ is the unit vector along y axis.

Write the expression for the vector form of force of magnitude 30.0N .

  F2=(30.0N)i^                                                                                                         (II)

Here, F2 is the force of magnitude 30.0N and i^ is the unit vector along x axis .

Write the expression for the vector form of force of magnitude 20.0N .

  F3=(20.0N)j^                                                                                                       (III)

Here, F3 is the force of magnitude 20.0N .

The force of magnitude 77.8N makes an angle 45° with x axis .

Write the expression for the vector form of force of magnitude 77.8N .

  F4=(77.8cos45°N)i^+(77.8sin45°N)j^=(55.01N)i^+(55.01N)j^                                                             (IV)

Here, F4 is the force of magnitude 77.8N .

Write the position vector of point where F1 is acting.

  r1=(2.0m)i^+(2.0m)j^                                                                                          (V)

Here, r1 is the position vector of point where F1 is acting.

Write the position vector of point where F2 is acting.

  r2=(2.0m)i^+(2.0m)j^                                                                                       (VI)

Here, r2 is the position vector of point where F2 is acting.

Write the position vector of point where F3 is acting.

  r3=(2.0m)i^+(2.0m)j^                                                                                      (VII)

Here, r3 is the position vector of point where F3 is acting.

Write the position vector of point where F4 is acting.

  r4=(2.0m)i^+(2.0m)j^                                                                                   (VIII)

Here, r4 is the position vector of point where F4 is acting.

Write general expression for r .

  r=rxi^+ryj^+rzk^                                                                                                   (IX)

Here, rx,ry, rz are the x,yand z component of r and i^ , j^ , k^ are unit vector along x, y, z directions.

Write the expression for torque.

  τ=r×F                                                                                                                  (X)

Here, τ is the torque on an object, r is the position vector of force from axis of rotation and F is the force acting on the object.

Write the general form of F .

  F=Fxi^+Fyj^+Fzk^                                                                                               (XI)

Here, Fx is the x component of force, Fy is the y component of force and Fz is the z component of force.

Substitute (IX) and (XI) in equation (X) to get τ .

  τ=(rxi^+ryj^+rzk^)×(Fxi^+Fyj^+Fzk^)=(ryFzrzFy)i^+(rzFxrxFz)j^+(rxFyryFx)k^                                               (XII)

Write the expression for the torque due to force F1 .

  τ1=r1×F1

Here, τ1 is the torque due to force F1 .

Write the expression for the torque due to force F2 .

  τ2=r2×F2

Here, τ2 is the torque due to force F2 .

Write the expression for the torque due to force F3 .

  τ3=r3×F3

Here, τ3 is the torque due to force F3 .

Write the expression for the torque due to force F4 .

  τ4=r4×F4

Here, τ4 is the torque due to force F4 .

Conclusion:

Substitute 0N for Fx, 60.0N for Fy and 0N for Fz, 2.0m for r1, 2.0m for r2 and 0.0m for r3 in (XII) to get τ1 .

  τ1=((2.0m)i^+(2.0m)j^+(0.0m)k^)×((0N)i^(60.0N)j^+(0N)k^)=((2.0m)(0N)(0.0m)(60.0N))i^+((0.0m)(0N)(2.0m)(0N))j^+((2.0m)(60.0N)(2.0m)(0N))k^=120k^Nm

Substitute 30N for Fx, 0N for Fy and 0N for Fz, 2.0m for r1, 2.0m for r2 and 0.0m for r3 in (XII) to get τ2 .

  τ2=((2.0m)i^+(2.0m)j^+(0.0m)k^)×((30N)i^+(0N)j^+(0N)k^)=((2.0m)(0N)(0.0m)(0N))i^+((0.0m)(30N)(2.0m)(0N))j^+((2.0m)(0N)(2.0m)(30N))k^=60k^Nm

Substitute 0N for Fx, 20N for Fy and 0N for Fz, 2.0m for r1, 2.0m for r2 and 0.0m for r3 in (XII) to get τ3 .

  τ3=((2.0m)i^+(2.0m)j^+(0.0m)k^)×((0N)i^+(20.0N)j^+(0N)k^)=((2.0m)(0N)(0.0m)(20.0N))i^+((0.0m)(0N)(2.0m)(0N))j^+((2.0m)(20.0N)(2.0m)(0N))k^=40k^Nm

Substitute 55.01N for Fx, 55.01N for Fy and 0N for Fz, 2.0m for r1, 2.0m for r2 and 0.0m for r3 in (XII) to get τ4 .

τ4=((2.0m)i^+(2.0m)j^+(0.0m)k^)×((55.01N)i^+(55.01N)j^+(0N)k^)=((2.0m)(0N)(0.0m)(55.01N))i^+((0.0m)(55.01N)(2.0m)(0N))j^+((2.0m)(55.01N)(2.0m)(55.01N))k^=(220k^)Nm

Add τ1, τ2, τ3 and τ4 to check whether it is zero not. If sum of torques are zero, the plate is in rotational equilibrium.

Write the expression for net torque.

  τ=τ1+τ2+τ3+τ4

Substitute 120k^Nm for τ1, 60k^Nm for τ2, 40k^Nm for τ3, 220k^Nm for τ4 in above equation to get τ .

  τ=120k^Nm+60k^Nm+40k^Nm+220k^Nm=0Nm

This indicates that the system is in rotational equilibrium.

Therefore, the pate is in equilibrium.

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Chapter 14 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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