Question

Review Three forces are applied to a wheel of radius 0.350 m, as shown in the figure (Figure 1). One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40.0° angle with the radius. Assume that Fi = 11.7N, F2 = 14.2N, and F3 = 8.80N Part A What is the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center? Indicate the direction of the net torque by letting counterclockwise torques be positive. Express your answer in newton meters to three significant figures. 10 AED ? Nm Submit Request Answer Provide Feedback Figure < 1 of 1 > 40.00 0.350 m

Answer 1

The torque due to the force F₁ is given by the relation,

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₁ = r x F = r F sin $\theta$
Where $\theta$ is the angle between the direction of force and the position vector r.
The force F₁ is acting to the centre of the wheel and thus the angle between force and the position vector is 180⁰.

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₁ = 0.
The torque acting on the wheel by the force F₂ is given by

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₂ = r x F₂ = r F₂ sin 40⁰ = 0.350 m x 14.2 N x sin 40

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₂ = - 3.195 Nm
This torque produces a clockwise rotation. And thus taken as negative.
The torque acting on the wheel by the force F₃

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₃ = r x F₃ = r F₃ sin 90⁰

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₃ = 0.350 m x 8.80 N x sin 90 = 3.08 N m.
This produces a counter-clockwise rotation and thus the torque is taken as positive.
Thus the net torque

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_net =
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₂ +
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₃ = - 3.195 Nm + 3.08 N m

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​​​​​​_net  = -0.115 N m