Question

cutnell & johnson problem 4.71
the drawing shows a large - cube (mass = 25 kg) being accelerated across a horizontal frictionless surface by a horizontal force $vec{p}$. a small cube (mass = 4.0 kg) is in contact with the front surface of the large cube and will slide downward unless $vec{p}$ is sufficiently large. the coefficient of static friction between the cubes is 0.71. what is the smallest magnitude that $vec{p}$ can have in order to keep the small cube from sliding downward?
frictionless $mu_{s}=0.71$

Explanation:

Step1: Analyze forces on small - cube

The maximum static - friction force \(f_s\) on the small cube must balance its weight \(mg\) to prevent it from sliding down. The formula for the maximum static - friction force is \(f_s=\mu_sN\), and we want \(f_s = mg\).
\[mg=\mu_sN\]
where \(m = 4.0\space kg\), \(g = 9.8\space m/s^2\), and \(\mu_s=0.71\).

Step2: Solve for the normal force \(N\)

From \(mg=\mu_sN\), we can solve for \(N\):
\[N=\frac{mg}{\mu_s}=\frac{4.0\times9.8}{0.71}\space N\]

Step3: Analyze the acceleration of the system

The normal force \(N\) on the small cube is provided by the acceleration of the system. The two - cube system has a total mass \(M = 25 + 4=29\space kg\). According to Newton's second law \(F = Ma\), and the acceleration \(a\) of the system is the same as the acceleration that causes the normal force on the small cube. Since \(N = ma\) (for the small cube), and \(F = Ma\) (for the two - cube system), we first find \(a=\frac{N}{m}\). Substituting \(N=\frac{mg}{\mu_s}\) into \(a=\frac{N}{m}\), we get \(a=\frac{g}{\mu_s}\).
Then, for the two - cube system with total mass \(M\), the force \(P\) is given by \(P = Ma\). Substituting \(a=\frac{g}{\mu_s}\) and \(M = 29\space kg\) into \(P = Ma\), we have:
\[P=\frac{M\times g}{\mu_s}\]
\[P=\frac{29\times9.8}{0.71}\]
\[P=\frac{284.2}{0.71}\approx400\space N\]

Answer:

\(400\space N\)