**Understanding Ideal Pulley Systems**To explore mechanical advantage (MA) through an ideal pulley system, let’s analyze the example below.**Problem Statement:**Suppose you use an ideal pulley, as illustrated, and find it necessary to exert a force of **240 N** to support a load.**Diagram Analysis:**- The diagram features a pulley system with a mechanical advantage (MA) of 3. - The force $ T $ is exerted through different sections of the rope and distributed across the pulleys.- The weight $ w $ is illustrated at the bottom, being supported by the pulley system.**Questions:**(a) **What is the load's mass (in kg)?**\[ \_\_\_\_ \text{ kg} \](b) **What force (in N) is exerted on the ceiling? Neglect the pulley's mass.**\[ \_\_\_\_ \text{ N (upward)} \]### Detailed Explanation:**Diagram Components:**- **Pulleys**: The system consists of multiple pulleys to distribute the weight, reducing the force required to lift the load.- **Tension (T)**: The diagram indicates $ T $ multiple times, showcasing the rope segments exerting tension.- **Mechanical Advantage (MA = 3)**: This suggests that the force exerted is magnified three times, allowing a lighter force to lift a heavier load.### Part (a) - Load's Mass Calculation:Given that the mechanical advantage (MA) is 3,\[ F_{\text{effort}} = 240 \text{ N} \]\[ MA = \frac{ \text{Load Force (F}_{\text{load}} \text{)}}{ \text{Effort force (F}_{\text{effort}} \text{)}} \]\[ 3 = \frac{ \text{Load Force}}{240 \text{ N}} \]\[ \text{Load Force} = 3 \times 240 \text{ N} \]\[ \text{Load Force} = 720 \text{ N} \]The load's weight is defined by:\[ w = mg \Rightarrow 720 \text{ N} = m \times 9.8 \text{ m/s}^2 \]\[ m = \frac{720 \text{ N}}{9.8 \

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Answered: 7) Suppose you use an ideal pulley of…

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7) Suppose you use an ideal pulley of the type shown in the figure below and find it necessary to exert a force of 240 N to support a load. MA-3 (a) What is the load's mass (in kg)? kg (b)What force (in N) is exerted on the ceiling? Neglect the pulley system's mass. N (upward)