Question

in the figure a crate of mass m = 75 kg is pushed at a constant speed up a frictionless ramp (θ=34) by a horizontal force F. The positive direction of an X-axis is up the ramp, and the positive direction of a y-axis is perpendicular to the ramp. (a) What is the magnitude of F? (b) What is the magnitude of the normal force on the crate?

Answer 1

a. |F| ≈ 496 N

b. normal force ≈ 887 N

Step-by-step explanation:

You want the magnitude of the horizontal force F that moves a crate up a 34° ramp at constant speed, and you want the magnitude of the normal force on the crate.

a) Force F

The constant speed of the crate tells you the net force up the ramp is zero. This is the sum of the component of force F in that direction and the force due to gravity in the opposite direction:

F·cos(34°) - m·g·sin(34°) = 0

F = mg·tan(34°) = (75 kg)(9.8 m/s²)tan(34°) ≈ 496 N

The magnitude of force F is about 496 N.

b) Normal force

The normal force on the crate will be the sum of the component of F in that direction and the force due to gravity in the same direction:

F·sin(34°) +m·g·cos(34°) ≈ 887 N

The magnitude of the normal force is about 887 N.

Answer 2

Step-by-step explanation:

Thanx for the figure;
The force component of F UP the ramp that moves the crate must equal the force of the crate DOWN the ramp

75 kg = mg Newtons = 735.8 Newtons

Downplane force is 735.8 sin 34° = 411.4 Newtons

Fn =The horizontal force will be found by cos 34 = 411.4/ F F = 411.4/cos (34) = 496 N

Normal Force = 735.8 cos 34° = 610 N This part is due to the mass of the crate....there is additional normal force from the force pushing the crate up the hill (from below)

= F sin34 = 496 sin 34 = 277.4 N

SUM of normal forces = 610 + 277.4 = 887.4 N