Question

16 mark for review a hanging block of mass 1.0 kg is attached by a string to a block of mass 2.0 kg that rests on a rough, horizontal surface, as shown in the figure. the mass of the string and pulley are negligible, and the pulley can rotate with negligible friction around its axle. what is the minimum coefficient of static friction $mu_s$ between the surface and the 2.0 kg block that would allow both blocks to remain at rest when released? a 0.33 b 0.50 c 0.67 d 1.00

Explanation:

Step1: Analyze forces on hanging block

The force due to gravity on the hanging block of mass $m_1 = 1.0\ kg$ is $F_{g1}=m_1g$. Using $g = 9.8\ m/s^{2}$, we have $F_{g1}=1.0\times9.8 = 9.8\ N$. This is the tension $T$ in the string when the system is on the verge of moving, so $T = m_1g$.

Step2: Analyze forces on horizontal - block

The normal force $N$ on the $m_2=2.0\ kg$ block resting on the horizontal surface is equal to its weight, so $N = m_2g$, where $m_2 = 2.0\ kg$ and $g = 9.8\ m/s^{2}$, then $N=2.0\times9.8 = 19.6\ N$. The maximum static - friction force $f_s=\mu_sN$ that can act on the $2.0\ kg$ block to prevent it from moving must balance the tension in the string. At the verge of motion, $f_s=T$.

Step3: Calculate coefficient of static friction

Since $f_s=\mu_sN$ and $f_s = T=m_1g$, $N = m_2g$, we can substitute to get $\mu_s=\frac{T}{N}=\frac{m_1g}{m_2g}=\frac{m_1}{m_2}$. Substituting $m_1 = 1.0\ kg$ and $m_2 = 2.0\ kg$, we have $\mu_s=\frac{1.0}{2.0}=0.50$.

Answer:

B. 0.50