The mechanism shown in figure is used to raise a wooden box of mass m = 50 kg. A string is wrapped around a cylinder that turns on an axle. The cylinder has radius 0.25 m and moment of inertia 1 kg m² about the axle. Magnitude of the force F applied tangentially to the rotating crank handle is required to raise the box with an acceleration of 2.0 m s⁻² is 1500 + 16a N. Here we can neglect the moment of inertia of the axle and the crank. Then a = ... (crank arm = 0.1 m).
Physics · Mixed Chapter
JEE Mains

Question

The mechanism shown in figure is used to raise a wooden box of mass m = 50 kg. A string is wrapped around a cylinder that turns on an axle. The cylinder has radius 0.25 m and moment of inertia 1 kg m² about the axle. Magnitude of the force F applied tangentially to the rotating crank handle is required to raise the box with an acceleration of 2.0 m s⁻² is 1500 + 16a N. Here we can neglect the moment of inertia of the axle and the crank. Then a = ... (crank arm = 0.1 m).

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The box mass is 50 kg accelerating upward at 2 m/s². With g=10, string tension is T=m(g+a)=50(12)=600 N. Step 2: The cylinder radius is r=0.25 m, so its angular acceleration is α=a/r=2/0.25=8 rad/s². Cylinder torque equation: F(0.1)−Tr=Iα. Step 3: F(0.1)=600(0.25)+(1)(8)=150+8=158 N·m. Thus F=1580 N. Step 4: The question writes F=1500+16a N; 1580=1500+16a ⇒ a=5.