Four forces are exerted on a disk of radius $ R $ that is free to spin about its center, as shown above. The magnitudes are proportional to the length of the force vectors, where $ F_1 = F_4 $, $ F_2 = F_3 $, and $ F_1 = 2F_2 $. Which two forces combine to exert zero net torque on the disk? • Nerd Notes

AP Physics

Unit 6 - Rotational Motion

Intermediate

Conceptual

MCQ

Four forces are exerted on a disk of radius $R$ that is free to spin about its center, as shown above. The magnitudes are proportional to the length of the force vectors, where $F_1 = F_4$ , $F_2 = F_3$ , and $F_1 = 2F_2$ . Which two forces combine to exert zero net torque on the disk?

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Step	Derivation/Formula	Reasoning
1	$\tau_1 = F_1 \times R$	The torque from force $F_1$ is calculated by the product of the force and the radius, acting perpendicular to the radius.
2	$\tau_2 = F_2 \times \frac{R}{2}$	Torque from force $F_2$ using its radius of action which is $\frac{R}{2}$ .
3	$\tau_3 = F_3 \times R$	Torque from force $F_3$ using full radius $R$ as it acts perpendicular.
4	$\tau_4 = F_4 \times R \times \sin(30^\circ)$	The torque from $F_4$ is calculated as it has a component $\sin(30^\circ)$ perpendicular to the radius.
5	$F_1 = 2F_2$	Given relationship, so $\tau_1 = 2F_2 R$ .
6	$\tau_1 = 4F_2 \times \frac{R}{2} = 4 \tau_2$	Substitute $F_1 = 2F_2$ and compare torques $\tau_1$ and $\tau_2$ .
7	$\tau_3 = F_2 \times R$	Use $F_3 = F_2$ , so torques are equal with opposite signs based on direction.
8	$\tau_4 = 0$	Not needed as it adds to $\tau_2$ but doesn’t balance others.
9	$\boxed{F_1, \ F_3}$	Forces $F_1$ and $F_3$ cancel each other because their torques are equal and opposite.

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Unit 6 - Rotational Motion

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Conceptual

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