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]
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]
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
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T1 = m (g + a)
T2 = m (g - a)
Apparent weight = mg – ma
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