Question

10.0-m long bar is attached by a frictionless hinge to a wall and held horizontal by a rope that makes an angle 53 degrees with the bar. The bar is uniform and weighs 39.9 N.

How far from the hinge should a 10.0-m Kg mass suspended from the tension T in the rope to be 125 N?

Answer 1

Torque = Force * perpendicular distance from the pivot point.

This is a torque problem
The weight of the 10 m bar is located 5 m from the hinge.(center of mass.
The weight of the 10 kg mass (98 N) is located d meters from the hinge.

Torque = Force * perpendicular distance from the pivot point.

These 2 forces pulling down on the bar, produce clockwise torque.

Clockwise torque = (39.9 N * 5 m) + (98 N * d)

The vertical component of the tension in the rope is located at the right end of the bar, 10 m from the hinge.

The vertical component of the tension in the rope = 125 N* sin 53