Let's solve the problem step by step: --- ### a) **Identify a coordinate system and draw free body diagrams for both blocks (6 pts)** Let’s choose: - For block \(2M\), the \(x\)-axis is horizontal to the right, and the \(y\)-axis is vertical upward. - For block \(M\), the \(y\)-axis is vertical (since it only moves up or down). - **Block \(2M\):** - Forces in play: - Applied force \(F\) at angle \(\theta\) to the left (decompose into \(F \cos\theta\) left, \(F\sin\theta\) up). - Tension \(T\) to the right (along the string). - Normal force \(N\) upward. - Gravitational force \(2Mg\) downward. - **Block \(M\):** - Forces in play: - Tension \(T\) upward. - Gravitational force \(Mg\) downward. **(You would draw these diagrams showing all forces as arrows)** --- ### b) **Write Newton's Second Law equations for both blocks (3 equations) (4 pts)** #### **For block \(2M\) (horizontal direction):** \[ \sum F_x = 2M a: \quad T - F\cos\theta = 2M a \] #### **For block \(2M\) (vertical direction):** \[ \sum F_y = 0: \quad N + F\sin\theta = 2Mg \] #### **For block \(M\) (vertical direction):** \[ \sum F_y = Ma: \quad Mg - T = M a \] (Note: \(a\) is the same for both blocks since they are connected by a string.) --- ### c) **Calculate the acceleration of the system, \(a\) (4 pts)** Add the two acceleration equations: 1. \(T - F\cos\theta = 2M a\) 2. \(Mg - T = M a\) Add (1) and (2): \[ T - F\cos\theta + Mg - T = 2M a + M a \] \[ Mg - F\cos\theta = 3M a \] \[ a = \frac{Mg - F\cos\theta}{3M} \] \[ \boxed{a = \frac{g - \frac{F\cos\theta}{M}}{3}} \] or, if you prefer: \[ \boxed{a = \frac{g - \frac{F\cos\theta}{M}}{3}} \] --- ### d) **Calculate the tension in the string, \(T\) (4 pts)** Use \(T - F\cos\theta = 2M a\): \[ T = 2M a + F\cos\theta \] Substitute \(a\) from above: \[ T = 2M \left( \frac{Mg - F\cos\theta}{3M} \right) + F\cos\theta \] \[ T = \frac{2(Mg - F\cos\theta)}{3} + F\cos\theta \] \[ T = \frac{2Mg - 2F\cos\theta}{3} + F\cos\theta \] \[ T = \frac{2Mg - 2F\cos\theta + 3F\cos\theta}{3} \] \[ T = \frac{2Mg + F\cos\theta}{3} \] \[ \boxed{T = \frac{2Mg + F\cos\theta}{3}} \] --- ### e) **What happens to acceleration if block \(2M\) is removed and the same force \(F\) is applied to block \(M\)? (2 pts)** Now, you have only block \(M\) hanging and being pulled directly by force \(F\): For block \(M\): \[ F\cos\theta + T = Mg \] But now, the force \(F\) is applied **directly** to the string supporting \(M\): So, the net force is \(Mg - F\cos\theta\) (downward), and the mass is \(M\): \[ a' = \frac{Mg - F\cos\theta}{M} = g - \frac{F\cos\theta}{M} \] **Compare to before:** Previously, \(a = \frac{Mg - F\cos\theta}{3M} = \frac{g - \frac{F\cos\theta}{M}}{3}\). So, **removing the block of mass \(2M\) increases the acceleration by a factor of 3**. **Explanation:** With only mass \(M\), there is much less inertia, so the same net force produces a much larger acceleration. --- ## **Summary of Final Answers** **a)** Coordinate system defined, FBDs drawn as described above. **b)** - \(T - F\cos\theta = 2M a\) - \(N + F\sin\theta = 2Mg\) - \(Mg - T = M a\) **c)** \(\boxed{a = \frac{g - \frac{F\cos\theta}{M}}{3}}\) **d)** \(\boxed{T = \frac{2Mg + F\cos\theta}{3}}\) **e)** The acceleration becomes \(a' = g - \frac{F\cos\theta}{M}\), which is **3 times larger** than before. With less mass, the same force causes more acceleration.