Answer:
The minimum downward force on the box that will keep it from slipping is 63.14 N
Explanation:
Given;
mass of the object, m = 30 kg
applied force, f = 125 N
coefficient of static friction, μ = 0.35
Normal reaction (R) is acting upwards, weight of the box (mg) is acting downwards and the minimum downward force (F) on the box that will keep it from slipping is also acting downwards.
The net vertical forces on the box is given by;
R - mg - F = 0
F = R - mg
Now, determine normal reaction, R
f = μR
R = f / μ
R = 125 / 0.35
R = 357.14 N
Finally, determine the minimum downward force on the box that will keep it from slipping;
F = R - mg
F = 357.14 - (30 x 9.8)
F = 357.14 - 294
F = 63.14 N
Therefore, the minimum downward force on the box that will keep it from slipping is 63.14 N
