The large quadriceps muscle in the upper leg terminates at its lower end in a tendon attached to the upper end of the tibia (see figure (a)). The forces on the lower leg when the leg is extended are modeled as in figure (b), where vector T is the force of tension in the tendon, vector w is the force of gravity acting on the lower leg, and vector F is the force of gravity acting on the foot. Find (the magnitude of) vector T when the tendon is at an angle θ of 25.0° with the tibia, assuming that w = 35.0 N, F = 12.4 N, and the leg is extended at an angle θ of 40.0° with the vertical. Assume that the center of gravity of the lower leg is at its center and that the tendon attaches to the lower leg at a point one-fifth of the way down the leg.
To find the magnitude of vector T, we need to dissect and analyze the forces acting on the lower leg.
Let's first consider the forces acting vertically. We have the force of gravity on the lower leg (w) and the force of gravity on the foot (F). These forces can be broken down into their vertical components:
wₓ = w * sin(θ)
Fₓ = F * sin(θ)
Now, let's consider the horizontal forces. The force of tension in the tendon (T) is the only horizontal force acting on the leg. The horizontal component of the force of gravity on the lower leg (w) will cancel out with the horizontal component of the force of gravity on the foot (F), so they do not contribute to the horizontal force.
The net horizontal force on the lower leg is equal to the tension in the tendon, T. We can find T by summing the horizontal forces:
T = wₓ + Fₓ
Substituting the expressions for wₓ and Fₓ, we have:
T = w * sin(θ) + F * sin(θ)
Now, let's plug in the given values:
w = 35.0 N
F = 12.4 N
θ = 25.0°
Substituting these values into the equation:
T = 35.0 N * sin(25.0°) + 12.4 N * sin(25.0°)
Calculating this expression, we find:
T ≈ 17.82 N
So, the magnitude of vector T is approximately 17.82 N when the tendon is at an angle of 25.0° with the tibia.