Cylinder rolling down an inclined plane held by a string

Question

A cylinder of mass M and radius R is in static equilibrium as shown in the diagram. The cylinder rests on an inclined plane making an angle with the horizontal and is held by a horizontal string attached to the top of the cylinder and to the inclined plane. There is friction between the cylinder and the plane. What is the tension in the string T?

Ans: $$T =Mg\sin(\theta)/(1 + \cos(\theta))$$

I'm having trouble arriving at this solution. I first look at the forces. I set my $x$-axis such that the Force of friction is parallel with the x-axis.

So my forces are:

$F_T$ = Force of Tension
$F_f$ = Force of friction
$F_N$ = Normal Force
$F_g$ = Force of Gravity

Since the cylinder is not moving. The forces and torques equal zero and balance.

Forces in the X direction:

$$0 = F_f + F_g\cos(\theta) - F_T*\cos(\theta)$$

Forces in the $y$ direction:

$$0 = F_N - F_g\sin(\theta) + F_T\sin(\theta) $$ $$F_T\sin(\theta) = F_g\sin(\theta) - F_N$$

$$F_T = (F_g\sin(\theta) - F_N)/\sin(\theta)$$

I'm not sure if I'm setting up my forces correctly. Can someone help?

Answer 1

If you equate the torque about the center of the cylinder you will find that only two force can produce torque. therefore:-
$$F_t*R=F_f*R$$, and the tension is equal to the friction. Then you write the equation in x direction along the slope of the incline :-
$$F_t*cos(\theta)+F_f-m*g*sin(\theta)=o $$
here you replace friction with tension and you will have the answer in required form.