Question

Three forces are exerted on the disk shown in the figure below, and their magnitudes are F3 = 2F, = 2F. The disk's outer rim has radius R, and the inner rim has radius R/2. As shown in the figure re tangent to the outer rim of the disk, a tangent to the inner rim. F3 is parallel to the x axis, F2 is parallel to the y axis, and Fi makes a 45° angle with the negative x axis. Find an expression for total torque exerted around the center of the disk in terms of R and F1. F ------

I have already tried (5*F₁*R)/2 and -(5*F₁*R)/2. None worked. Not sure what I did wrong. Please Help.

Answer 1

The net torque on the disk is

$\tau=-F_1\times R-F_3\times R+F_2\times (R/2)$

We have

$F_3=2F_2=2F_1$

$\Rightarrow F_2=F_1$

And

$\Rightarrow F_3=2F_1$

That gives us

$\tau=-F_1\times R-2F_1\times R+F_1\times (R/2)$

$\tau=-3F_1R+\frac{F_1R}{2}=\frac{-6F_1R+F_1R}{2}=-\frac{5F_1R}{2}$

The direction of the torque is in the z-direction. So, the final answer is

$\vec \tau=-\frac{5F_1R}{2}\hat k$

or

$\vec \tau=-\frac{5F_1R}{2}\hat z$

NOTE: Since the answer is asked in the form of a vector. You need to provide a direction in the answer.