Question

a 12 kg sphere is supported by two ropes ab and bc. what is the tension force in rope ab and bc?
a 10 kg block is placed on top of the 20 kg block as shown. the system of two blocks can move on a horizontal frictionless surface. the coefficient of static friction between the blocks is 0.6. if a horizontal 150 n force is applied to the 20 kg block and the top block doesnt slide on the bottom block, what is the magnitude of the horizontal force that the 10 kg block exerts on the 20 kg block?
based on the graph above, a 10 kg block is placed on top of the 20 kg block as shown. the system of two blocks can move on a horizontal frictionless surface. the coefficient of static friction between the blocks is 0.6. if a horizontal 150 n force is applied to the 20 kg block and the top block doesnt slide on the bottom block, what is the acceleration of the system?

Explanation:

Step1: Analyze forces on 12 - kg sphere

The weight of the 12 - kg sphere is $F_g=mg$, where $m = 12\ kg$ and $g=9.8\ m/s^2$. So $F_g=12\times9.8 = 117.6\ N$. Let the tension in rope AB be $T_{AB}$ and in rope BC be $T_{BC}$. In the vertical - direction, $T_{BC}\sin37^{\circ}=F_g$. In the horizontal - direction, $T_{AB}=T_{BC}\cos37^{\circ}$.
First, from $T_{BC}\sin37^{\circ}=F_g$, we can solve for $T_{BC}$:
$T_{BC}=\frac{F_g}{\sin37^{\circ}}$. Since $\sin37^{\circ}\approx0.6$, $T_{BC}=\frac{117.6}{0.6}=196\ N$.

Step2: Calculate $T_{AB}$

Using $T_{AB}=T_{BC}\cos37^{\circ}$, and $\cos37^{\circ}\approx0.8$, we have $T_{AB}=196\times0.8 = 156.8\ N$.

Step3: Analyze the two - block system

For the two - block system (total mass $m_{total}=10 + 20=30\ kg$), according to Newton's second law $F = m_{total}a$, where $F = 150\ N$.
$a=\frac{F}{m_{total}}=\frac{150}{30}=5\ m/s^2$.
The force that the 10 - kg block exerts on the 20 - kg block and the force that the 20 - kg block exerts on the 10 - kg block are a pair of action - reaction forces.
The maximum static friction force between the blocks is $f_{s,\max}=\mu_sN$, where $N = m_{10}g=10\times9.8 = 98\ N$, and $f_{s,\max}=0.6\times98 = 58.8\ N$.
The acceleration of the 10 - kg block is the same as the acceleration of the system (since it doesn't slide), and the force on the 10 - kg block is $F_{10}=m_{10}a=10\times5 = 50\ N$. According to Newton's third law, the force that the 10 - kg block exerts on the 20 - kg block has a magnitude of $50\ N$.

Answer:

• Tension in rope AB: $156.8\ N$
• Tension in rope BC: $196\ N$
• Force that 10 - kg block exerts on 20 - kg block: $50\ N$
• Acceleration of the system: $5\ m/s^2$