The femur of a human leg (mass 10 kg, length 0.9 m) is in traction, as shown in the figure below. The center of gravity of the leg is one-third of the distance from the pelvis to the bottom of the foot. Two objects, with masses m1 and m2, are hung using pulleys to provide upward support. A third object of 8 kg is hung to provide tension along the leg. The body provides tension as well. What is the mathematical relationship between m1 and m2? Is this relationship unique in the sense that there is only one combination of m1 and m2 that maintains the leg in static equilibrium?
The mathematical relationship between m1 and m2 is m1 + m2 = 8 kg. This relationship is not unique, as there are multiple combinations of m1 and m2 that can be used to maintain the leg in static equilibrium. For example, m1 could be 4 kg and m2 could be 4 kg, or m1 could be 6 kg and m2 could be 2 kg.
To find the mathematical relationship between m1 and m2 and determine if it is unique, we need to analyze the forces acting on the leg in static equilibrium.
Let's break down the problem step by step:
1. Center of Gravity (CoG): The center of gravity is located one-third of the distance from the pelvis to the bottom of the foot. This information is important because it helps establish the distribution of weight along the leg.
2. Traction Force: The tension being applied to the leg comes from two sources - the third object (8 kg) and the body itself. Let's call the tension from the third object T1 and the tension from the body T2.
3. Upward Support: There are two objects, m1 and m2, hanging from pulleys that provide upward support. The tension in the ropes attached to m1 and m2 will counterbalance the combined weight of the leg (10 kg + 8 kg) and the tension provided by the body.
Now, let's consider the forces acting on the leg:
1. Weight of the Leg: The weight of the leg acts vertically downward. Its value is given as the mass (10 kg) multiplied by the acceleration due to gravity (9.8 m/s^2). Therefore, the weight of the leg is 10 kg * 9.8 m/s^2 = 98 N.
2. Tension from the Third Object (T1): The third object (8 kg) hanging on the leg provides tension along the leg. This tension force acts vertically upward.
3. Tension from the Body (T2): The body also applies tension to the leg, acting vertically upward.
4. Tension from m1 and m2 (Tension in Ropes): The ropes attached to m1 and m2 provide upward support. Both tensions act vertically upward.
Now, let's analyze the equilibrium conditions:
1. Vertical Equilibrium:
In static equilibrium, the sum of all the vertical forces must be zero. Therefore:
Sum of upward forces = Weight of the Leg + T1 + T2 + Tension from m1 + Tension from m2
2. Distribution of Weight:
Since the center of gravity is symmetrical, the tension from m1 and m2 should be distributed equally. Thus, the tension from m1 = the tension from m2.
Now, assuming Tension from m1 = Tension from m2 = T, we can rewrite the vertical equilibrium equation:
Weight of the Leg + T1 + T2 + 2T = 0
From the given information in the problem, we can find that T1 = 8 kg * 9.8 m/s^2 (weight of the third object), and T2 is the tension exerted by the body.
In conclusion, the mathematical relationship between m1 and m2 is that they should have the same magnitude of tension (T). This relationship is unique as long as the tensions T1 and T2 are correctly determined based on the given information.
To determine the mathematical relationship between m1 and m2, let's analyze the forces acting on the femur and the system as a whole.
1. Center of Gravity of the Leg:
The center of gravity of the leg is one-third of the distance from the pelvis to the bottom of the foot. This means the distance from the pelvis to the center of gravity of the leg is (1/3) * length = (1/3) * 0.9 m = 0.3 m.
2. Forces acting on the femur:
- Weight of the femur (Wf): The weight of the femur is given by Wf = m * g, where m = 10 kg is the mass of the femur and g = 9.8 m/s^2 is the acceleration due to gravity. Therefore, Wf = 10 kg * 9.8 m/s^2 = 98 N.
- Tension from the body (Tb): The body provides tension to counteract the downward force of the femur. Let Tb be the tension exerted by the body.
- Tension from the third object (T3): The third object with a mass of 8 kg provides tension along the leg. Let T3 be the tension provided by the third object.
3. Forces acting on the system:
- Tension from object m1 (T1): The object m1 provides upward tension support. Let T1 be the tension provided by m1.
- Tension from object m2 (T2): The object m2 also provides upward tension support. Let T2 be the tension provided by m2.
- Total upward tension (Tu): The total upward tension in the system is the sum of the tensions from m1 and m2. Tu = T1 + T2.
4. Static Equilibrium Condition:
For the femur to be in static equilibrium, the sum of all the forces acting on it must be zero, and the sum of all the torques acting on it must also be zero.
- Sum of forces in the vertical direction:
The sum of the vertical forces is given by:
Tb + T3 + T1 + T2 - Wf = 0
- Sum of torques about the center of gravity:
The torque due to the femur weight is equal to the weight multiplied by the distance between the center of gravity and the point of rotation (pelvis):
(0.3 m) * Wf = (0.3 m) * (10 kg * 9.8 m/s^2)
The torque due to T3 is equal to T3 multiplied by the distance between the tension force and the point of rotation (pelvis), which is 0.3 - 0.3 = 0 m:
0.3 m * T3 = 0
The torques due to T1 and T2 can be neglected since their distances to the center of gravity are not given.
5. Mathematical Relationship between m1 and m2:
Since we know that T1 + T2 = Tu, we can rewrite the first equilibrium equation from step 4 in terms of Tu:
Tb + T3 + Tu - Wf = 0
Substituting Tu = T1 + T2, we have:
Tb + T3 + T1 + T2 - Wf = 0
Simplifying the equation, we get:
Tb + T3 = Wf - T1 - T2
Because T1 + T2 = Tu, we can substitute Tu in the equation:
Tb + T3 = Wf - Tu
Therefore, the mathematical relationship between m1 and m2 is:
m1 + m2 = Wf / g - m3 (where m3 is the mass of the third object, m3 = 8 kg)
6. Uniqueness of the Relationship:
From the above equation, there can be multiple combinations of m1 and m2 that satisfy the equation and maintain the leg in static equilibrium. Therefore, the relationship is not unique in the sense that there is more than one combination of m1 and m2 that can achieve static equilibrium.