The coefficient of static friction between the box and the ramp is 0.500. What is the magnitude of the minimum force, F, that must be applied to the box perpendicularly to the ramp to prevent the box from sliding?

31.86N

Explanation:

The question is incomplete. Here is the complete question.

"A 58 kg box is at rest on a ramp, which is inclined 28° above the horizontal.The coefficient of static friction between the box and the ramp is 0.500. What is the magnitude of the minimum force, F, that must be applied to the box perpendicularly to the ramp to prevent the box from sliding?"

Given the following data;

Mass of the box = 58kg

Angle of inclination θ =28°

coefficient of static friction between the box and the ramp μ = 0.500

The forces acting on the box on the ramp is as shown in the attachment.

It can be seen that the normal reaction force N and the force perpendicular to the ramp Fp preventing the body from sliding acts in the same direction.

The net reaction force acting on the body can be expressed as;

Rt = N + Fp

According to the diagram,

N = W = mgcosθ

Rt = mgcosθ + Fp ... (1)

To prevent the box from sliding, the frictional force Ff acting on the body alond the plane must be equal to the force Fm causing the body to slide down the plane (moving force).

Mathematically;

Fm = Ff (for static body)

Ff= μRt ... (2)

Substituting Rt in equation 1 into equation 2 to get the net frictional force;

Ff = μ(mgcosθ + Fp) ... (3)

Also Fm = mgsinθ ... (4)

Equating equation 3 and 4 to get Fp we have;

μ(mgcosθ + Fp) = mgsinθ

Substituting the values given in the resulting equation we have;

0.5{58(9.81)cos28° + Fp} = 58(9.81)sin28°

0.5{568.98cos28° + Fp} = 568.98sin28°

0.5{502.38 +Fp} = 267.12

251.19+0.5Fp = 267.12

0.5Fp = 267.12-251.19

0.5Fp = 15.93

Fp = 15.93/0.5

Fp = 31.86N

The force perpendicular to the ramp required to prevent the box from sliding is 31.86N.

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