A block of mass m is placed on another block of mass M which itself is lying on a horizontal surface. The coefficient of friction between the block of mass M and horizontal surface is μ$$1$$ and that between the two blocks is μ$$2$$ . What maximum horizontal force can be applied to the lower block so that the two blocks move without separation?

Question

A block of mass m is placed on another block of mass M which itself is lying on a horizontal surface. The coefficient of friction between the block of mass M and horizontal surface is μ$$1$$ and that between the two blocks is μ$$2$$ . What maximum horizontal force can be applied to the lower block so that the two blocks move without separation?

Infinity Learn · Accepted Answer

Maximum acceleration of m by friction $$=$$ their maximum common acceleration μ$$2mgm=F$$−μ$$1(M+m)gM+msolving$$ we get,$$F=(M+m)($$μ$$2+$$μ$$1)g$$

Two block problem

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Maximum acceleration of m by friction = their maximum common acceleration
$\begin{array}{l} \frac{μ_{2} mg}{m} = \frac{F - μ_{1} (M + m) g}{M + m} \\ solving we get, \\ F = (M + m) (μ_{2} + μ_{1}) g \end{array}$

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Two block problem

Remember concepts with our Masterclasses.

Maximum acceleration of m by friction = their maximum common acceleration μ2mgm=F−μ1(M+m)gM+msolving we get,F=(M+m)(μ2+μ1)g

Ready to Test Your Skills?

free study material

Maximum acceleration of m by friction = their maximum common acceleration
$\begin{array}{l} \frac{μ_{2} mg}{m} = \frac{F - μ_{1} (M + m) g}{M + m} \\ solving we get, \\ F = (M + m) (μ_{2} + μ_{1}) g \end{array}$