Question

3) Weightlifting Weightlifting can exert extremely large forces on the human body, especially on joints and tendons. Consider the strict curl, a lift in which the lifter uses both arms to lift a barbell. The lift is done with a pivot of the elbow. The biceps muscle contracts, pulling up on the forearm through the tension in a tendon. The weight is lifted using only the forearms, with the upper arms remaining stationary. The world record for the strict curl is above 90.0 kg. The figure shows a simplified version of this lift. Biceps Muscle Upper Arm Forearm Tendon Barbell 4.00 cm 35.0 om , a) Suppose a 91.77 kg barbell is held stationary with the forearms horizontal. Find the magnitude of the upward tension in the tendon that connects the biceps muscle to the bone in each arm. (Assume the two arms contribute equally to the lift.) Hint: Model the two portions of the arm as two rods connected by a hinge at the elbow. The mass of the forearm itself is small compared to the mass of the barbell, and it can be ignored. b) Find the magnitude and direction of the force exerted on the forearm at the elbow.

Answer 1

Solution

Data

W = 91.77kg

r1 = 4.00cm = 0.04m

r2 = 35.0cm = 0.35m

To solve this problem, we perform the following scheme

Biceps muscle Tendon 14.00 cm Elbow Free- body diagram 35.0 cm The 4.00 cm 35.0 cm

Where:

W: is the weight of the bar

T: is the  upward tension in the bicepts

F_E: is the force exerted on the elbow

a) The magnitud of the upward tension is calculated, as follows:

Point A( pivot at the elbow)

(1)

ΣΤΑ = 0

In reference to the free body diagram, we have:

(2)

Fe*0+ (0.047) - (0.35 * 91.77kg * 9.8m/s?) = 0

As the mass of the forearm is much smaller than the mass of the bar, its torque is not considerable

The torque of the force passing on the elbow is zero, since its line of action passes through that point of support.

Then, the tension force

- (0.35 # 91.77kg * 9.8m/s2) – 7.87.10°N 0.04

b) In this case, we apply as follows:

(3)

F =0

(4)

T - FE-W =0

Then

FE=T-W

Fe = 7.87 . 10°N - (91.77kg 9.8m/s) N = 6.97. 10%N

The direction the FA is $270^{\circ}$ with respect the axes horizontal.