2) The uniform beam to the right has a mass of 65kg and a total length of 3m. The beam is supported by a pivot point on its far-left end. Determine the upward force, F, needed to balance the beam. This force is located 2.45m from the pivot point . This image contains three physics problems regarding torque, force, and angular displacement.**1) Pulley Torque Problem:**- **Problem Statement:** Refer to the image on the right. If \( M_1 = 4.5 \, \text{kg} \) and \( M_2 = 10.2 \, \text{kg} \), determine the total torque on the pulley. The radius of the pulley is 13 cm. - **Diagram Description:** A pulley system is shown with two masses, \( M_1 \) and \( M_2 \), hanging on either side of the pulley. The radius \( R \) is marked on the pulley.**2) Beam Force Problem:**- **Problem Statement:** The uniform beam to the right has a mass of 65 kg and a total length of 3 m. The beam is supported by a pivot point on its far-left end. Determine the upward force, \( F \), needed to balance the beam. This force is located 2.45 m from the pivot point. - **Diagram Description:** A beam is shown with a pivot on its left end and an upward force \( F \) applied near the right end.**3) Angular Displacement Problem:**- **Problem Statement:** A shaft turning at 750 rad/s slows to a stop in 6.55 seconds. What is the angular displacement the shaft rotates during this time?These problems involve calculating physical quantities using physics principles such as torque, force balance, and rotational kinematics. The accompanying diagrams illustrate the scenarios described.

Given values: Mass of the beam, m=65 kg Length of the beam, L=3m Distance between pivot and force point, D=2.45 mThe forces acting on the beam are , Force exerted by the pivot( Vertical and horizontal component) Force due to gravity(Downward) Applied force( Upward) At equilbrium, Net force, Fnet=0 Net torque, τnet=0 Torque is the product force and perpendicular distance. τ=rFsinθ Where, r=Distance from the axis of rotation to the point of application of forceF=Forceθ=Angle between r and F Torques on the beam are, Torque due to both the components of pivot force are zero. Because ,r=0 Torque due to force of gravity, τg=L2mgsin90° (Clockwise) Torque due to applied force, τF=DFsin90°(Anti-clockwise) Applying equilibrium condition of torque, τg-τF=0L2mgsin90°-DFsin90°=0L2mgsin90°=DFsin90°mgL2=DF (1) Acceleration due to gravity, g=9.8 m/s2 Found the value of applied force by substituting the values of mass of the beam, acceleration due to gravity, Length of the beam, distance between the force point and pivot point in equation (1). 65 kg×9.8 m/s2×3 m2=2.45 m×FF=65 kg×9.8 m/s2×3 m2×2.45 m=390 N Answer: The upward force balances the beam is 390 N