Question

A large cube (mass = 25 kg) being accelerated across a horizontal frictionless surface by a horizontal force . A small cube mass = 4.0 kg) is in contact with the front surface of the large cube and will slide downward unless is sufficiently large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that can have in order to keep the small cube from sliding downward?

Answer 1

Answer:F=400.28 N

Explanation:

Given

mass of large cube $M=25 kg$

mass of small cube $m=4 kg$

coefficient of static friction $\mu =0.71$

acceleration a of the system is given by

$F=(M+m)a$

$a=\frac{F}{M+m}$

Now Normal reaction between two blocks is given by

$N=ma$

friction tries to balance weight

thus $f_r=mg$

and $f_r=\mu N$

$f_r=\mu ma$

$\mu ma=mg$

$\mu \frac{F}{M+m}=g$

$F=\frac{(M+m)g}{\mu }$

$F=\frac{29\times 9.8}{0.71}$

$F=400.28 N$