A18.4-kg box rests on a frictionless ramp with a 16.1° slope. the mover pulls on a rope attached to the box to pull it up the incline. if the rope makes an angle of 43.2° with the horizontal, what is the smallest force f the mover will have to exert to move the box up the ramp?

F= 56,1 N :

Explanation:

We apply Newton's first law for balancing forces system.

We take the x axis in the direction of the ramp with a 16.1° slope.

∑Fx= 0

∑Fx: algebraic sum of forces ( + to the right, - to the left)  

Problem development:

Look at the free body diagram :

The only forces that act on the box are the weight and tension of the rope because there is no friction.

T: Rope tension (N)

W :Box weight (N)

∑Fx= 0

Tx-Wx=0

Tx =Tcos( 43.2°- 16.1°)= Tcos ( 27.1°)

Wx= W*sen  16.1°= m*g*sen  16.1°=18.4*9.8* sen  16.1° = 50N

Tcos ( 27.1°)-50=0

Tcos ( 27.1°) = 50

T = (50) / (cos ( 27.1°))

T = 56,1 N

T=F= 56,1 N

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement