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27 P

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Physics For Scientists & Engineers
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Short Answer

Question: Find the net torque on the wheel in Figure P10.27 about the axle through a=10.0cm, taking and b=25.0cm.

The net torque on the wheel about the axle through O taking a=10.0cm and b=25.0cm is τ=3.71.

See the step by step solution

Step by Step Solution

Step 1: Defining torque

The force that makes an object rotate is a torque.

τ=rFsinθ

Step 2: Calculating the net torque

Consider the given figure.

Find the torque for 10.0N.

τ10N=rF

Since the radius of the torque 10N is b.

Substitute r = 0.25m, F=10N

τ10N=0.25×10

τ10N=2.5Nm

Thus, the torque rotates in clockwise direction.

Calculate the torque for 9.00N.

Let the radius of 9.00N is b,

Substitute, r = 0.25m,

F = 9N

Therefore,

τ9N=rF

τ9N=0.25×9

τ9N=2.25N, rotates in clockwise direction.

Determine the torque for 12.0N

Since it has an angle, the force is perpendicular.

Thus, the radius is from the figure,

Substitute the following values:

r=0.1m,F=12N,θ=30°

Therefore

τ12N=12cos30°×0.1τ12N=12×32×0.1τ12N=12×0.866×0.1τ12N=1.0392Nm

It rotates in anticlockwise direction.

Hence, the total torque can be calculated as

τ=τ10N+τ9N-τ12Nτ=2.5Nm+2.25Nm-1.039Nmτ=3.71Nm

Therefore, the net torque exerted on the wheel is τ=3.71Nm.

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