Two blocks (m = 0.5kg and M = 4.5kg) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is 3/7. Then the maximum horizontal force that can be applied on the block so that blocks move together is ---------N. (Round off to the Nearest Integer) [ Take g as 9.8 ms^-2]

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Answered by

Vishal Baghel | Contributor-Level 10

a month ago

For the combined system of mass M and m, the acceleration under an applied force F is:
a = F / (M + m)

The static friction force (f_s) on the top block (m) provides its acceleration:
f_s = MA = m * [F / (M + m)] = mF / (M + m)

For the top block not to slip, the required static friction must be less than or equal to the maximum possible static friction (μmg):
f_s ≤ μmg
mF / (M + m) ≤ μmg
F ≤ μ (M + m)g

Using the values implied in the solution:
F ≤ 21 N

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