torque help

[pizza1994]

Advertisement - Members don't see this ad

Assume that a massless bar 5 meters in length is suspended from a rope and that the rope is attached to the bar at a distance x from the bar’s left end. If a 20-kg mass hangs from the right side of a bar and a 5-kg mass hangs from the left side of the bar, what value of x will bring about equilibrium?

The answer is 4 m.

can someone help me solve this question using the center of mass equation where the LHS is the pivot point?

Thanks!!!

 

[sillyjoe]

So in this problem you are looking for the pivot point for this system to bring about equilibrium i.e. the point at which the rope can be hung so that the counterclockwise torque will equal the clockwise torque.

torque = rF(sin(theta)). Since the scenario described has the masses hanging perpendicular to the system you can drop sin(theta) out of the equation since sin(90) = 1.

Let X = pivot point, which in this case is the rope

rFcounterclockwise = rFclockwise

(5kg)(10m/s^2)(x) = (20kg)(10m/s^2)(5-x)

x = 4 meters