The force exerted on the strut by the hinge if the strut is weightless.

Question

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Answer 1

Question

Chapter 12, Problem 23P

(a)

To determine

To Find:The force exerted on the strut by the hinge if the strut is weightless.

(a)

Expert Solution

Answer to Problem 23P

F→=(30.0N)i^+(30.0N)j^

Explanation of Solution

Given data:

The strut is weightless.

Weight of the block, 60 N

Formula used:

Static translational equilibrium condition:

∑F→=0

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 12, Problem 23P

The force exerted on the strut at the hinge:

F→=Fhi^+fvj^ (1)

Ignore the weight of the strut and use ∑τ→=0 at the hinge:

LT−(Lcos45∘)W=0

T=Wcos45∘=(60 N) cos45∘ =42.43 N

Use ∑F→=0 for the strut:

∑Fx=Fh−Tcos45∘=0

∑Fy=Fv+Tcos45∘−Mg=0

Solve for Fh :

Th=Tcos45∘=(42.43 N)cos45∘ =30 N

Solve for Fv :

Fv=Mg−Tcos45∘ =60N−(42.43N)cos45∘ =30.0 N

Subtitute in equation (1) to obtain:

F→=(30.0N)i^+(30.0N)j^

Conclusion:

F→=(30.0N)i^+(30.0N)j^

(b)

To determine

The force exerted on the strut by the hinge if the strut weighs 20 N .

(b)

Expert Solution

Answer to Problem 23P

F→=(35.0N)i^+(45.0N)j^

Explanation of Solution

Given data:

Strut weighs, 20 N

Weight of the block, 60 N

Formula used:

Translational and rotational equilibrium:

∑F=0∑τ=0

Calculation:

Including the weight of the strut, apply ∑τ→=0

LT−(Lcos45∘)W−(L2cos45)w=0

T=(cos45∘)W+(12cos45∘)w =(cos45∘)(60N)+(12cos45∘)(20N) =49.5 N

Apply ∑F→=0 to the strut:

∑Fx=Fh−Tcos45∘=0

And

∑Fy=Fv+Tcos45∘−W−w=0

Solve for Fh :

Th=Tcos45∘=(49.5N)cos45∘ =35.0N

Solver for Fv :

Fv=W+w−Tcos5∘ =60 N+20 N-(49.5 N)cos45∘ =45.02 N

Substitute in equation (1) to obtain:

F→=(35.0N)i^+(45.0N)j^

Conclusion:

F→=(35.0N)i^+(45.0N)j^

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Answer 2

Question

Chapter 12, Problem 23P

(a)

To determine

To Find:The force exerted on the strut by the hinge if the strut is weightless.

(a)

Expert Solution

Answer to Problem 23P

F→=(30.0N)i^+(30.0N)j^

Explanation of Solution

Given data:

The strut is weightless.

Weight of the block, 60 N

Formula used:

Static translational equilibrium condition:

∑F→=0

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 12, Problem 23P

The force exerted on the strut at the hinge:

F→=Fhi^+fvj^ (1)

Ignore the weight of the strut and use ∑τ→=0 at the hinge:

LT−(Lcos45∘)W=0

T=Wcos45∘=(60 N) cos45∘ =42.43 N

Use ∑F→=0 for the strut:

∑Fx=Fh−Tcos45∘=0

∑Fy=Fv+Tcos45∘−Mg=0

Solve for Fh :

Th=Tcos45∘=(42.43 N)cos45∘ =30 N

Solve for Fv :

Fv=Mg−Tcos45∘ =60N−(42.43N)cos45∘ =30.0 N

Subtitute in equation (1) to obtain:

F→=(30.0N)i^+(30.0N)j^

Conclusion:

F→=(30.0N)i^+(30.0N)j^

(b)

To determine

The force exerted on the strut by the hinge if the strut weighs 20 N .

(b)

Expert Solution

Answer to Problem 23P

F→=(35.0N)i^+(45.0N)j^

Explanation of Solution

Given data:

Strut weighs, 20 N

Weight of the block, 60 N

Formula used:

Translational and rotational equilibrium:

∑F=0∑τ=0

Calculation:

Including the weight of the strut, apply ∑τ→=0

LT−(Lcos45∘)W−(L2cos45)w=0

T=(cos45∘)W+(12cos45∘)w =(cos45∘)(60N)+(12cos45∘)(20N) =49.5 N

Apply ∑F→=0 to the strut:

∑Fx=Fh−Tcos45∘=0

And

∑Fy=Fv+Tcos45∘−W−w=0

Solve for Fh :

Th=Tcos45∘=(49.5N)cos45∘ =35.0N

Solver for Fv :

Fv=W+w−Tcos5∘ =60 N+20 N-(49.5 N)cos45∘ =45.02 N

Substitute in equation (1) to obtain:

F→=(35.0N)i^+(45.0N)j^

Conclusion:

F→=(35.0N)i^+(45.0N)j^

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Answer 3

Question

Chapter 12, Problem 23P

(a)

To determine

To Find:The force exerted on the strut by the hinge if the strut is weightless.

(a)

Expert Solution

Answer to Problem 23P

F→=(30.0N)i^+(30.0N)j^

Explanation of Solution

Given data:

The strut is weightless.

Weight of the block, 60 N

Formula used:

Static translational equilibrium condition:

∑F→=0

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 12, Problem 23P

The force exerted on the strut at the hinge:

F→=Fhi^+fvj^ (1)

Ignore the weight of the strut and use ∑τ→=0 at the hinge:

LT−(Lcos45∘)W=0

T=Wcos45∘=(60 N) cos45∘ =42.43 N

Use ∑F→=0 for the strut:

∑Fx=Fh−Tcos45∘=0

∑Fy=Fv+Tcos45∘−Mg=0

Solve for Fh :

Th=Tcos45∘=(42.43 N)cos45∘ =30 N

Solve for Fv :

Fv=Mg−Tcos45∘ =60N−(42.43N)cos45∘ =30.0 N

Subtitute in equation (1) to obtain:

F→=(30.0N)i^+(30.0N)j^

Conclusion:

F→=(30.0N)i^+(30.0N)j^

(b)

To determine

The force exerted on the strut by the hinge if the strut weighs 20 N .

(b)

Expert Solution

Answer to Problem 23P

F→=(35.0N)i^+(45.0N)j^

Explanation of Solution

Given data:

Strut weighs, 20 N

Weight of the block, 60 N

Formula used:

Translational and rotational equilibrium:

∑F=0∑τ=0

Calculation:

Including the weight of the strut, apply ∑τ→=0

LT−(Lcos45∘)W−(L2cos45)w=0

T=(cos45∘)W+(12cos45∘)w =(cos45∘)(60N)+(12cos45∘)(20N) =49.5 N

Apply ∑F→=0 to the strut:

∑Fx=Fh−Tcos45∘=0

And

∑Fy=Fv+Tcos45∘−W−w=0

Solve for Fh :

Th=Tcos45∘=(49.5N)cos45∘ =35.0N

Solver for Fv :

Fv=W+w−Tcos5∘ =60 N+20 N-(49.5 N)cos45∘ =45.02 N

Substitute in equation (1) to obtain:

F→=(35.0N)i^+(45.0N)j^

Conclusion:

F→=(35.0N)i^+(45.0N)j^

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Answer 4

Question

Chapter 12, Problem 23P

(a)

To determine

To Find:The force exerted on the strut by the hinge if the strut is weightless.

(a)

Expert Solution

Answer to Problem 23P

F→=(30.0N)i^+(30.0N)j^

Explanation of Solution

Given data:

The strut is weightless.

Weight of the block, 60 N

Formula used:

Static translational equilibrium condition:

∑F→=0

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 12, Problem 23P

The force exerted on the strut at the hinge:

F→=Fhi^+fvj^ (1)

Ignore the weight of the strut and use ∑τ→=0 at the hinge:

LT−(Lcos45∘)W=0

T=Wcos45∘=(60 N) cos45∘ =42.43 N

Use ∑F→=0 for the strut:

∑Fx=Fh−Tcos45∘=0

∑Fy=Fv+Tcos45∘−Mg=0

Solve for Fh :

Th=Tcos45∘=(42.43 N)cos45∘ =30 N

Solve for Fv :

Fv=Mg−Tcos45∘ =60N−(42.43N)cos45∘ =30.0 N

Subtitute in equation (1) to obtain:

F→=(30.0N)i^+(30.0N)j^

Conclusion:

F→=(30.0N)i^+(30.0N)j^

(b)

To determine

The force exerted on the strut by the hinge if the strut weighs 20 N .

(b)

Expert Solution

Answer to Problem 23P

F→=(35.0N)i^+(45.0N)j^

Explanation of Solution

Given data:

Strut weighs, 20 N

Weight of the block, 60 N

Formula used:

Translational and rotational equilibrium:

∑F=0∑τ=0

Calculation:

Including the weight of the strut, apply ∑τ→=0

LT−(Lcos45∘)W−(L2cos45)w=0

T=(cos45∘)W+(12cos45∘)w =(cos45∘)(60N)+(12cos45∘)(20N) =49.5 N

Apply ∑F→=0 to the strut:

∑Fx=Fh−Tcos45∘=0

And

∑Fy=Fv+Tcos45∘−W−w=0

Solve for Fh :

Th=Tcos45∘=(49.5N)cos45∘ =35.0N

Solver for Fv :

Fv=W+w−Tcos5∘ =60 N+20 N-(49.5 N)cos45∘ =45.02 N

Substitute in equation (1) to obtain:

F→=(35.0N)i^+(45.0N)j^

Conclusion:

F→=(35.0N)i^+(45.0N)j^

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Answer 5

Chapter 12, Problem 23P

(a)

To determine

To Find:The force exerted on the strut by the hinge if the strut is weightless.

(a)

Expert Solution

Answer to Problem 23P

F→=(30.0N)i^+(30.0N)j^

Explanation of Solution

Given data:

The strut is weightless.

Weight of the block, 60 N

Formula used:

Static translational equilibrium condition:

∑F→=0

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 12, Problem 23P

The force exerted on the strut at the hinge:

F→=Fhi^+fvj^ (1)

Ignore the weight of the strut and use ∑τ→=0 at the hinge:

LT−(Lcos45∘)W=0

T=Wcos45∘=(60 N) cos45∘ =42.43 N

Use ∑F→=0 for the strut:

∑Fx=Fh−Tcos45∘=0

∑Fy=Fv+Tcos45∘−Mg=0

Solve for Fh :

Th=Tcos45∘=(42.43 N)cos45∘ =30 N

Solve for Fv :

Fv=Mg−Tcos45∘ =60N−(42.43N)cos45∘ =30.0 N

Subtitute in equation (1) to obtain:

F→=(30.0N)i^+(30.0N)j^

Conclusion:

F→=(30.0N)i^+(30.0N)j^

(b)

To determine

The force exerted on the strut by the hinge if the strut weighs 20 N .

(b)

Expert Solution

Answer to Problem 23P

F→=(35.0N)i^+(45.0N)j^

Explanation of Solution

Given data:

Strut weighs, 20 N

Weight of the block, 60 N

Formula used:

Translational and rotational equilibrium:

∑F=0∑τ=0

Calculation:

Including the weight of the strut, apply ∑τ→=0

LT−(Lcos45∘)W−(L2cos45)w=0

T=(cos45∘)W+(12cos45∘)w =(cos45∘)(60N)+(12cos45∘)(20N) =49.5 N

Apply ∑F→=0 to the strut:

∑Fx=Fh−Tcos45∘=0

And

∑Fy=Fv+Tcos45∘−W−w=0

Solve for Fh :

Th=Tcos45∘=(49.5N)cos45∘ =35.0N

Solver for Fv :

Fv=W+w−Tcos5∘ =60 N+20 N-(49.5 N)cos45∘ =45.02 N

Substitute in equation (1) to obtain:

F→=(35.0N)i^+(45.0N)j^

Conclusion:

F→=(35.0N)i^+(45.0N)j^

Answer 6

Chapter 12, Problem 23P

(a)

To determine

To Find:The force exerted on the strut by the hinge if the strut is weightless.

(a)

Expert Solution

Answer to Problem 23P

F→=(30.0N)i^+(30.0N)j^

Explanation of Solution

Given data:

The strut is weightless.

Weight of the block, 60 N

Formula used:

Static translational equilibrium condition:

∑F→=0

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 12, Problem 23P

The force exerted on the strut at the hinge:

F→=Fhi^+fvj^ (1)

Ignore the weight of the strut and use ∑τ→=0 at the hinge:

LT−(Lcos45∘)W=0

T=Wcos45∘=(60 N) cos45∘ =42.43 N

Use ∑F→=0 for the strut:

∑Fx=Fh−Tcos45∘=0

∑Fy=Fv+Tcos45∘−Mg=0

Solve for Fh :

Th=Tcos45∘=(42.43 N)cos45∘ =30 N

Solve for Fv :

Fv=Mg−Tcos45∘ =60N−(42.43N)cos45∘ =30.0 N

Subtitute in equation (1) to obtain:

F→=(30.0N)i^+(30.0N)j^

Conclusion:

F→=(30.0N)i^+(30.0N)j^

Answer 7

Chapter 12, Problem 23P

(a)

To determine

To Find:The force exerted on the strut by the hinge if the strut is weightless.

Answer 8

(a)

Expert Solution

Answer to Problem 23P

F→=(30.0N)i^+(30.0N)j^

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given data:

The strut is weightless.

Weight of the block, 60 N

Formula used:

Static translational equilibrium condition:

∑F→=0

Calculation:

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 12, Problem 23P

The force exerted on the strut at the hinge:

F→=Fhi^+fvj^ (1)

Ignore the weight of the strut and use ∑τ→=0 at the hinge:

LT−(Lcos45∘)W=0

T=Wcos45∘=(60 N) cos45∘ =42.43 N

Use ∑F→=0 for the strut:

∑Fx=Fh−Tcos45∘=0

∑Fy=Fv+Tcos45∘−Mg=0

Solve for Fh :

Th=Tcos45∘=(42.43 N)cos45∘ =30 N

Solve for Fv :

Fv=Mg−Tcos45∘ =60N−(42.43N)cos45∘ =30.0 N

Subtitute in equation (1) to obtain:

F→=(30.0N)i^+(30.0N)j^

Conclusion:

F→=(30.0N)i^+(30.0N)j^

Answer 13

(b)

To determine

The force exerted on the strut by the hinge if the strut weighs 20 N .

(b)

Expert Solution

Answer to Problem 23P

F→=(35.0N)i^+(45.0N)j^

Explanation of Solution

Given data:

Strut weighs, 20 N

Weight of the block, 60 N

Formula used:

Translational and rotational equilibrium:

∑F=0∑τ=0

Calculation:

Including the weight of the strut, apply ∑τ→=0

LT−(Lcos45∘)W−(L2cos45)w=0

T=(cos45∘)W+(12cos45∘)w =(cos45∘)(60N)+(12cos45∘)(20N) =49.5 N

Apply ∑F→=0 to the strut:

∑Fx=Fh−Tcos45∘=0

And

∑Fy=Fv+Tcos45∘−W−w=0

Solve for Fh :

Th=Tcos45∘=(49.5N)cos45∘ =35.0N

Solver for Fv :

Fv=W+w−Tcos5∘ =60 N+20 N-(49.5 N)cos45∘ =45.02 N

Substitute in equation (1) to obtain:

F→=(35.0N)i^+(45.0N)j^

Conclusion:

F→=(35.0N)i^+(45.0N)j^

Answer 14

(b)

To determine

The force exerted on the strut by the hinge if the strut weighs 20 N .

Answer 15

(b)

Expert Solution

Answer to Problem 23P

F→=(35.0N)i^+(45.0N)j^

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given data:

Strut weighs, 20 N

Weight of the block, 60 N

Formula used:

Translational and rotational equilibrium:

∑F=0∑τ=0

Calculation:

Including the weight of the strut, apply ∑τ→=0

LT−(Lcos45∘)W−(L2cos45)w=0

T=(cos45∘)W+(12cos45∘)w =(cos45∘)(60N)+(12cos45∘)(20N) =49.5 N

Apply ∑F→=0 to the strut:

∑Fx=Fh−Tcos45∘=0

And

∑Fy=Fv+Tcos45∘−W−w=0

Solve for Fh :

Th=Tcos45∘=(49.5N)cos45∘ =35.0N

Solver for Fv :

Fv=W+w−Tcos5∘ =60 N+20 N-(49.5 N)cos45∘ =45.02 N

Substitute in equation (1) to obtain:

F→=(35.0N)i^+(45.0N)j^

Conclusion:

F→=(35.0N)i^+(45.0N)j^

Answer 20

Ch. 12 - Prob. 1P Ch. 12 - Prob. 2P Ch. 12 - Prob. 3P Ch. 12 - Prob. 4P Ch. 12 - Prob. 5P Ch. 12 - Prob. 6P Ch. 12 - Prob. 7P Ch. 12 - Prob. 8P Ch. 12 - Prob. 9P Ch. 12 - Prob. 10P

The force exerted on the strut by the hinge if the strut is weightless.

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To Find:The force exerted on the strut by the hinge if the strut is weightless.

Answer to Problem 23P

Explanation of Solution

The force exerted on the strut by the hinge if the strut weighs 20 N .

Answer to Problem 23P

Explanation of Solution

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