A person is trying to drag a box with a mass of 30 kg on a flat floor using a rope attached to it. The static frictional coefficient between the box and floor is 0.50. The angle between the horizontal direction and the rope is 30 degrees. What is the minimum force the person needs to have to move the box along the floor?

F = 131.7N

Explanation:

You are given the following parameters.

Mass M = 30 kg

Coefficient of static friction μ = 0.5

Ø = 30 degrees

When the person is trying to drag the box with force F, the static frictional force Fs will be acting in the opposite direction.

From the figure attached, resolve all forces into horizontal component and vertical component.

Horizontal component:

Fs - F cosØ = 0

Fs = F cosØ

F cosØ = μN ...... (1)

Vertical component:

N + F SinØ - mg = 0

N = Mg - F SinØ ..... (2)

Substitutes m, g and Ø into the equation 2

N = (30 × 9.8) - F × sin30

N = 294 - 0.5F

Substitute N and coefficient of friction into the equation (1)

F cos30 = 0.5 (294 - 0.5F)

Open the bracket

0.8660F = 147 - 0.25F

Collect the like terms

1.116025F = 147

F = 147/1.116025

F = 131.7 N

Therefore, the minimum force the person needs to have to move the box along the floor is 131.7 N

The answer is 4f======since electrostatic force is given by    we let f be the electrostatic force when  r = 1 m    if we let  denote the electrostatic force when the distance is decreased so that r = 0.5m, then    we established earlier that f = kq₁q₂. therefore    the force will now be 4f.

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