A 18.4-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 i
ncline. 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?
1 answer:
Answer:
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
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