Question

In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm 3.80 cm from the elbow joint. Assume that the person's hand and forearm together weigh 15.0 N and that their center of gravity is 15.0 cm from the elbow (not quite halfway to the hand). The forearm is held horizontally at a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the forearm.
A. Find the force exerted by the biceps when the hand is empty.
B. Now the person holds a 80.0-N weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is 33.0 cm from the elbow. Find the force now exerted by the biceps.
C. Explain why the biceps muscle needs to be very strong.
D. Under the conditions of part B, find the magnitude of the force that the elbow joint exerts on the forearm.
E. Under the conditions of part B, find the direction of the force that the elbow joint exerts on the forearm.
F. While holding the 80.0-N weight, the person raises his forearm until it is at an angle of 53.0∘ above the horizontal. If the biceps muscle continues to exert its force perpendicular to the forearm, what is this force when the forearm is in this position?
G. Has the force increased or decreased from its value in part B? Explain why this is so, and test your answer by actually doing this with your own arm.

Answer 1

Answer:

Answer is explained in the explanation section below.

Explanation:

Part A)

From conserve moment of force, we have:

F1d1 = F2d2

F1 x (3.80 x $10^{-2}$ m) = 15N x (15 x $10^{-2}$ m)

F1 = $\frac{15 . 15 . 10^{-2} }{3.80 . 10^{-2} }$

F1 = 59.2 N

Force exerted by the biceps when the hand is empty.

Part B)

The 80 N weight acts at 33 cm and 15 N at 15 cm, then the center of mass is:

x = $\frac{m1x1 + m2x2}{m1+m2}$

x = $\frac{\frac{80}{9.8} (33 .10^{-2}) + \frac{15}{9.8}(15.10^{-2} }{\frac{80}{9.8} + \frac{15}{9.8} }$

x = 30.16 x $10^{-2}$ m

Total Weight is:

F = 80N + 15N = 95N

From the conserve moment of force, we have:

F ( 3.8 x $10^{-2}$ ) = 95N (30.16 x $10^{-2}$)

F = 754 N

Part C:

From the above two examples solved, the force exerted by the biceps is higher than downward force, due to this muscle need to be very strong.

Part D)

The force exerted by elbow on the forearm is:

The force exerted by the elbow and biceps are in upward direction and total weight is in downward direction. So, the balancing force in vertical direction is:

F2 + 754N = 95N

F2 = 95N -754N

F2 = -659N

Negative sign shows the force is in downward direction.

Part E)

The bicep muscle acts perpendicular to the forearm, so it is lever arm stays the same. but those of the other two forces decreases as the arm is raised. There tension in the biceps muscle decreases.

Part F)

Angle = 53 degrees.

So,

Force = FcosФ

Force = 754 cos 53

Force = 453.76 N

Part G)

The value of force has gone downwards. It has decreased from that of part B.