Question

Two blocks, stacked one on top of the other, slide on a frictionless horizontal surface. The surface between the two blocks is rough, however, with a coefficient of static friction equal to 0.40. The top block has a mass of 2.9 kg, and the bottom block's mass is 5.8 kg. If a horizontal force F is applied to the bottom block, what is the maximum value F can have before the top block begins to slip?

Answer 1

Answer:

The value is $F = 11.37 \ N$

Explanation:

From the question we are told that

    The coefficient of static friction is  $\mu_s = 0.40$

    The mass of the top of block  is  $m_t = 2.9 \ kg$

     The mass of the  bottom is  $m_b = 5.8 \ kg$

Generally given that the frictional force between the bottom block and the frictionless horizontal is zero then the maximum value F can have before the top block begins to slip will be equal to the frictional force between the top and  the bottom block which is mathematically represented as

      $F = F_f =\mu_s * N$

Here N is the normal force acting on top block which is mathematically represented as

     $N = m_t * g$

    $N = 2.9 * 9.8$

=>$N = 28.42 \ N$

So

     $F = 0.40 * 28.42$

=>   $F = 11.37 \ N$