A 23 kg suitcase is being pulled with a constant speed by a handle that is at an angle of 29° above the horizontal. If the normal force exerted on the suitcase is 160 n, what is the force f applied to the handle?

The force F applied to the handle = 330.03 N

Explanation:

The force can be resolved in to two, horizontal component and vertical component. If θ is the angle between horizontal and applied force we have

Horizontal component of force = F cos θ

Vertical component of force = F sin θ

In this problem normal force exerted on the suitcase is 160 N, that is vertical component of force = 160 N and angle θ = 29⁰.

So, F sin 29 = 160

F = 330.03 N

The force F applied to the handle = 330.03 N

329.89N

Explanation:

Normal force is a force that is perpendicular to the horizontal surface. The normal force will therefore acts in the positive y direction.

To get the force F applied at the handle, we will resolve the force to the component of the normal force (y component)

The component of the force along the positive y axis (Fy) will be + Fsin (theta) where theta is 29°

Since both the normal force and the resolved force are acting in the same direction, Fy = normal force

Fsin (theta) = 160N

Fsin29° = 160N

F = 160/sin29°

F = 160/0.485

F = 329.89N

Therefore the force F applied to the handle is 329.89N