Answer to Question #155145 in Mechanics | Relativity for NISHANT KUMAR MEENA

Question

 Stacked Blocks F т, m2 no friction Consider two blocks that are resting one on top of the other. The lower block has mass = 2.2 frictionless table. The upper block has mass m kg. Suppose the coefficient of static friction between the two blocks is given by 4.8 kg and is resting m2= on a = 0.5 A force of magnitude F is applied as shown. What is Fmaz, the maximum value of F for which the upper block can be pushed horizontally together without slipping? Please give your answer in Newtons. so that the two blocks move

Answer 1

Explanations & Calculations

• Consider the figure attached

• Since all the boxes are on a frictionless table, for any value F the system gets accelerated due to the friction between the blocks.
• That is caused by the forward frictional force $\small f$ on $\small m_2$
• Even though the system moves the 2 blocks remain in equilibrium so that $\small F=f$ on the upper block.
• Then the $\small F$ can be increased until the static frictional conditions are met which is $\small F_{max}=f_{max}=\mu_sR$
• Any value for $\small F$ beyond this causes the upper block to slip on the other block.
• The normal reaction $R$ on the upper block can be calculated considering the vertical equilibrium of the upper block, that is $\small \uparrow R-m_1g=0\to R=m_1g$
• Then the maximum value for the applied force is

$\qquad\qquad \begin{aligned} \small F_{max}&= \small \mu_s (m_1g) \end{aligned}$

• The numerical values are not very clear so that a numerical answer cannot be gained but algebraically solved sot that you can plug the relevant numerical values.