This is my problem:
A block of mass 1.50 kg is accelerated across a rough surface by a light cord passing over a small pulley shown in the image below. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.140 m above the top of the block. The coefficient of kinetic friction is 0.360.
This is the question it is asking:
Find the largest value of x for which the acceleration is zero.
I have been trying to solve this for hours with no luck, please help.
Edit: To further explain my problem and what I have tried:
I know that to find x when acceleration = 0 you set the force of friction equation equal to force giving you Tcos(theta) = mu(mg - Tsin(theta) and then you put it in terms of x. I got (1/m)(mu)(m)(g)[x^2 + h^2]^-1/2 = (1/m)Tx + (1/m)(mu)(h)(T) (h being the height of the pulley). I then substitute in my variables m = 1.5kg, h = 0.14m, mu = 0.36, and T = 10N. I guess what I’m asking is if that equation is correct because it worked for a practice problem with different numbers but not the numbers in the problem where it counts.
You have
$T\cos(\theta) = \mu (mg-T\sin(\theta))$
$\cos(\theta) = \frac{x}{\sqrt{x^2+h^2}}$
$\sin(\theta) = \frac{h}{\sqrt{x^2+h^2}}$
plugging these into the original equation we get:
$T\frac{x}{\sqrt{x^2+h^2}} = \mu (mg-T\frac{h}{\sqrt{x^2+h^2}})$
multiply both sides by $\sqrt{x^2+h^2}$
Is this what you also got?
$Tx = \mu (mg\sqrt{x^2+h^2}-Th)$
Now plug in everything and solve for x. You'll have to isolate the square root. Then square both sides, simplify and use the quadratic formula. What do you get?