Question

An overhead door is guided by wheels at a and b that roll in horizontal and vertical tracks. when θ = 40°, the velocity of wheel b is 1.8 ft/s upward. determine the angular velocity of the door and the velocity of end d of the door.

Answer 1

I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door $d$)...

The door forms a right triangles that satisfies

$\tan\theta=\dfrac ab\implies\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{b\frac{\mathrm da}{\mathrm dt}-a\frac{\mathrm db}{\mathrm dt}}{b^2}$

We also have

$\tan\theta=\dfrac ab\implies\cos\theta=\dfrac bd$

so if you happen to know the height of the door, you can solve for $b$ and $a$.

$d$ is fixed, so

$a^2+b^2=d^2\implies2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}=0\implies\dfrac{\mathrm da}{\mathrm dt}=-\dfrac ba\dfrac{\mathrm db}{\mathrm dt}$

We can solve for the angular velocity $\dfrac{\mathrm d\theta}{\mathrm dt}$:

$\dfrac{\mathrm d\theta}{\mathrm dt}=\cos^2\theta\dfrac{b\left(-\frac ba\frac{\mathrm db}{\mathrm dt}\right)-a\frac{\mathrm db}{\mathrm dt}}{b^2}=-\dfrac1a\dfrac{\mathrm db}{\mathrm dt}$

At the point when $\theta=40^\circ$ and $\dfrac{\mathrm db}{\mathrm dt}=1.8$ ft/s, we get

$\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{1.8}a\dfrac{\rm deg}{\rm s}=-\dfrac{1.8}{d\sin40^\circ}\dfrac{\rm deg}{\rm s}$