Question

A 1130-kg car is held in place by a light cable on a smooth (frictionless) ramp. The cable makes an angle of 31.0° above the surface of the ramp, and the ramp itself rises at 25.0° above the horizontal. Determine the tension in the cable?

Answer 1

Answer:

T = 5163.89 N

Explanation:

Newton's first law:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

We define the x-axis in the direction parallel to the movement of the car on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the car

W: Weight of the car : In vertical direction

FN : Normal force : perpendicular to the ramp

T :Tension force:  at angle of 31.0° above the surface of the ramp

Calculated of the Weight  of the car (W)

W = m*g   m: mass   g:acceleration due to gravity

W =   1130-kg* 9.8 m/s² = 11074 N

x-y weight components

Wx =  11074 N*sin 25.0° = 4680.07 N

Wy = 11074 N*cos 25.0° = 10036.45 N

x-y Tension components

Tx = T*cos 25.0°

Ty = T*sin 25.0°

Newton's first law:

∑Fx =0 Formula (1)

Tx-Wx = 0

T*cos 25.0° - 4680.07 = 0

T*cos 25.0° = 4680.07

T =  4680.07 / cos 25.0°

T = 5163.89 N