a standard man intending to do reverse curls in a gym. He holds his arms straight, using an overhand grip to hold the bar. If the mass of the bar is given as 140 kg, what is the tension in each of his shoulders? Consider the weight of the arm and the weight of the bar. Note: Use 4.9kg as the mass of the arm of the standard man.

Question

a standard man intending to do reverse curls in a gym. He holds his arms straight, using an overhand grip to hold the bar. If the mass of the bar is given as 140 kg, what is the tension in each of his shoulders? Consider the weight of the arm and the weight of the bar. Note: Use 4.9kg as the mass of the arm of the standard man.

Answer 1

I don't do reverse curls, and am not too sure what it is. This is what I learned:

http://www.exrx.net/WeightExercises/Brachioradialis/BBReverseCurl.html

I suppose the best approximation is to assume that the two shoulders assume the total weight of the bar and the arms (4.9 kg includes the wrist and the fingers, hopefully).

So load each shoulder would shoulder
(140+4.9+4.9)/2=74.9 kg = 734 N

However, since it is a tension on the arm, I am not too sure if it would be a tension on the shoulder. The calculation of the muscle/tendon forces are rather complex without a good knowledge of human anatomy.

Answer 2

To find the tension in each of the man's shoulders, we need to consider the weight of both the bar and the man's arms.

The force exerted by the bar can be calculated using the formula F = m * g, where F is the force, m is the mass, and g is the acceleration due to gravity.

The mass of the bar is given as 140 kg. The acceleration due to gravity is approximately 9.8 m/s^2.

So, the force exerted by the bar is F_bar = 140 kg * 9.8 m/s^2.

Next, let's calculate the force exerted by the man's arms. The weight of an object is given by the formula W = m * g.

The mass of the man's arm is given as 4.9 kg. So, the weight of one arm is W_arm = 4.9 kg * 9.8 m/s^2.

Since the man is using both arms, the total weight of his arms is 2 * W_arm.

Now, the tension in each of the man's shoulders is given by the sum of the forces exerted by the bar and his arms. Therefore, the tension T in each shoulder is T = F_bar + 2 * W_arm.

Substituting the given values, T = (140 kg * 9.8 m/s^2) + 2 * (4.9 kg * 9.8 m/s^2).

Now, let's calculate the tension:

T = (1372 N) + 2 * (48 N) = 1372 N + 96 N = 1468 N.

Therefore, the tension in each of the man's shoulders is 1468 Newtons.