Problems

Q1

The biceps muscle exerts a vertical force of 700N on the lower arm as shown below. Assume that the muscle is attached 5cm from the elbow. Calculate the torque about the axis of rotation through the elbow joint due to the weight of the lower arm and hand ?

Sol:

Given the arm is at angle (45°) below the horizontal.

Therefore the angle between the force

and the position vector

(r = 5cm) is 45°

Now we need to find the lever arm

, inorder to calculate the torque.

The arm can exert less torque at this angle than when it is at 90° with the force.

Q2

The drawing shows a lower leg being exercised. It has a 40N weight attached to the foot and is extended at an angle θ with respect to the vertical. Consider a rotational axis at the knee

(a) Find the magnitude of torque that weight creates when θ = 90°

(b) At what angle θ does the magnitude of torque equal 25N - m?

Sol:

Weight attached to the leg is F = 40N

Distance from the axis (Knee) to the weight r = 0.65m

(a) Given that the foot is extended at an angle θ = 90° with respect to the vertical.

In this case we need not find the lever arm,

the given distance is itself the lever arm, as the force is acting perpendicularly.

(The leg is perpendicular to the knee, which is the axis)

Free body diagram

(b) Given magnitude of torque τ = 25N - m

From the figure, the angle between the force

and the position vector

is also θ(r is a transversal intersecting two parallel lines)

So at 74° the magnitude of torque is equal to 25 N-m

Q3

One end of a rod is hinged to a table, so that the rod can rotate freely on the table top.Two forces, both parallel to the table top, act on the rod at the same place. One force is directed perpendicular to the rod and has a magnitude of 30N. The second force has a magnitude of 50N and is directed at an angle θ with respect to the rod. Tf the sum of the torques due to the two forces is zero, what must be the angle θ?

Sol: