In the drawing below, the leg of a child is in cast and supported in a traction device. The system is in equilibrium and both pulleys in the diagram are frictionless. The mass of the ropes are negligible. Determine the magnitude of the force F exerted by the leg on the small pulley. HINT: First find the tension in the cable.
Let's call the tension in the cable T. Since the system is in equilibrium, the net force on the leg must be zero.
First, let's look at the forces acting on the leg. There is the force of gravity pulling the leg downward, with a magnitude of m*g, where m is the mass of the leg and g is the acceleration due to gravity (9.8 m/s^2). There is also a force T acting on the leg horizontally, pulling it to the right.
Now let's look at the forces acting on the small pulley. There is tension T pulling it to the right, and there is an equal and opposite tension T pulling it to the left from the other side of the cable.
Since the pulley is stationary, the net force on it must be zero. Therefore, the tension in the cable is equal to the force exerted by the leg on the pulley (F), which is equal to T. So we just need to find T.
Looking at the forces acting on the leg, we can set up the following equation:
T - m*g = 0
Solving for T:
T = m*g
Now we can substitute in the values given in the problem. The mass of the leg is 20 kg, so:
T = 20 kg * 9.8 m/s^2 = 196 N
Therefore, the magnitude of the force F exerted by the leg on the small pulley is also 196 N.