Question

Most of the motion generated by joints in the human body are examples of levers. An example of a third class lever is the forearm; the force which generates the motion is located in between the pivot point and the resistance force. The elbow acts as the pivot point, the biceps muscle pulls up on the forearm to move the mass which is held in the hand. Imagine you are holding an object in your hand with your forearm horizontal and your upper arm vertical so your elbow creates a 90° angle. If your biceps attach to your forearm 0.0640 m from the elbow and make an angle of 28.0° with the vertical while a 2.38 kg mass is held in your hand 28.60 cm from the elbow, what force is exerted by the biceps muscle? Ignore the mass of the forearm.

Answer 1

Answer 2

The force applied by the biceps can be found by summing the torques about the elbow joint. Take a clockwise rotation to be positive.

$$\sum \left(\tau \right)=0=F\cos \left(\theta \right)d-\left(mg{d}_1\right)$$

Here, ?$\theta$ is the angle the biceps makes with the vertical, d?d is the distance of the bicep from the elbow,  m is the mass in the hand, and $d_1$ is the distance from the center of the mass in the hand to the elbow. Solve this equation for the force applied by the biceps. Make sure to convert units into standard SI.

$$F=\frac{mg{d}_1}{\cos \left(\theta \right)d}$$

Substitute the given quantities.

?=2.00 kg⋅9.81 m/s2⋅(28.60 cm⋅1 m100 cm)cos(29.0∘)⋅(0.0640 m)=1.00×102 N