Question

If a 150 pound weight is on a frictionless surface, raised at an angle of 35 degrees, what is the tension in the rope that keeps it from sliding down? What is the force perpendicular to the surface?

Answer 1

This is the given situation.

Where m is the mass of the block and g is the acceleration due to gravity. It is given in the question, mg=150 pound=68.04 kg.

There are two components of weight. One along with x-direction and the other with negative y-direction.

x-component is

`mg\sin \theta`

y-component is

`mg\cos \theta`

Tension on the string is equal to the x-component of the weight. and the normal force,i.e. perpendicular force is equal and opposite to the y component of the weight. But tension is in opposite direction to the x-component of weight and perpendicular force is opposite to the y-component.

Therefore the tension is,

`T=-mg\sin \theta=-68.04*\sin 35^o=-39.03\text{ N}`

And the normal force is,

`N=mg\cos \theta=60.04*\cos 35^o=55.74\text{ N}`

Therefore the magnitude of the tension on the string is 39.03 N

And the normal force is 55.74 N