A variation Russel traction supports the lower leg in a cast. suppose that the patient's leg and cast have a combined mass of 15.0kg and m1 is 4.50kg. (a) what is reaction force of the leg muscles to the traction? (b) what must m2 be to keep the leg horizontal?
A figure must go with this question. You need to do a force and torque balance.
To find the reaction force of the leg muscles to the traction (a), we need to remember that the sum of all forces acting on an object in equilibrium is equal to zero.
(a) Let's consider the forces acting on the system.
1. The weight of the leg and cast (combined mass of 15.0 kg) acts downward, which can be represented as m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
2. The tension in the traction device, which is supporting the leg, acts upward. We need to find this tension force.
Now, since the leg is in equilibrium, the sum of the forces is zero. Therefore, we can write the equation:
Tension force - Weight of the leg and cast = 0
Tension force = Weight of the leg and cast
Tension force = m1 * g
Substituting the given values, we have:
Tension force = 4.50 kg * 9.8 m/s²
Calculate the values to find the tension force.
(b) In order to keep the leg horizontal, the tension force in the traction (Tension force) should equal the weight of the leg and cast (Weight of the leg and cast), given by m1 * g.
Since the leg needs to be horizontal, there should not be any unbalanced force on it. In this case, the reaction force exerted by the traction device is equal to the weight of the leg and cast. Therefore, to keep the leg horizontal, m2 should be equal to m1, which is 4.50 kg in this case.