NLM BYU 21

Step 1: Analyze the System Logic

For block B and ball C to remain motionless relative to block A, the entire system (A + B + C) must accelerate horizontally with a common acceleration $a$. The problem states there is no contact between A and C. This implies that the ball C is hanging freely from the pulley, supported only by the tension $T$, and it maintains its position relative to A due to the system’s acceleration.

Step 2: Free Body Diagram of C (in the frame of A)

Since C is motionless relative to A, we can resolve forces in the accelerated frame. The string makes an angle $\theta$ with the vertical.

• Vertical: Tension component balances gravity: $T \cos \theta = m_C g$
• Horizontal: Tension component provides the necessary acceleration (or balances pseudo force): $T \sin \theta = m_C a$

From these two equations, the total tension $T$ is the vector sum of $m_C g$ and $m_C a$:

$$T = \sqrt{(m_C g)^2 + (m_C a)^2} = m_C \sqrt{g^2 + a^2}$$

Step 3: Free Body Diagram of B

Block B moves horizontally with acceleration $a$. Since the surface is frictionless, the only horizontal force acting on B is the tension $T$ from the string.

$$T = m_B a$$

Step 4: Solving for Acceleration

Equating the expressions for Tension $T$:

$$m_B a = m_C \sqrt{g^2 + a^2}$$

Squaring both sides:

$$m_B^2 a^2 = m_C^2 (g^2 + a^2)$$ $$m_B^2 a^2 – m_C^2 a^2 = m_C^2 g^2$$ $$a^2 (m_B^2 – m_C^2) = m_C^2 g^2 \implies a = \frac{m_C g}{\sqrt{m_B^2 – m_C^2}}$$

Step 5: Total Force F

The external force $F$ pushes the total mass of the system $(m_A + m_B + m_C)$ with acceleration $a$.

$$F = (m_A + m_B + m_C) a$$

Substituting the value of $a$:

$$F = \frac{(m_A + m_B + m_C) m_C g}{\sqrt{m_B^2 – m_C^2}}$$