Question

Block A, mass 250 g , sits on top of block B, mass 2.0 kg . The coefficients of static and kinetic friction between blocks A and B are 0.34 and 0.23, respectively. Block B sits on a frictionless surface. What is the maximum horizontal force that can be applied to block B, without block A slipping

Answer 1

Answer:

  F = 69.3 N

Explanation:

For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by

               fr = μ N

We define a reference system parallel to the floor

block B  ( lower)

Y axis  

            N - W₁-W₂ = 0

            N = W₂ + W₂

            N = (M + m) g

X axis

              F -fr = M a

for block A (upper)

X axis

              fr = m a                 (2)

so that the blocks do not slide, the acceleration in both must be the same.

Let's solve the system by adding the two equations

             F = (M + m) a          (3)

             a =$\frac{F}{ M+m}$

the friction force has the formula

            fr = μ N

             fr = μ (M + m) g

let's calculate

            fr = 0.34 (2.0 + 0.250) 9.8

            fr = 7.7 N

we substitute in equation 2

             fr = m a

             a = fr / m

             a = 7.7 / 0.250

             a = 30.8 m / s²

we substitute in equation 3

             F = (2.0 + 0.250) 30.8

             F = 69.3 N