Question

The bones of the forearm (radius and ulna) are hinged to the humerus at the elbow. The biceps muscle connects to the bones of the forearm about 2.152.15 cm beyond the joint. Assume the forearm has a mass of 2.452.45 kg and a length of 0.4650.465 m. When the humerus and the biceps are nearly vertical and the forearm is horizontal, if a person wishes to hold an object of mass 6.556.55 kg so that her forearm remains motionless, what is the force exerted by the biceps muscle?

Answer 1

Answer:

$F_b$  = 1647.92 N

Explanation:

Given:

length of the arm, r = 0.465 m

distance of forearm from elbow, r' = 2.15 cm = 0.0215 m

Mass of the forearm, M = 2.45 kg

Mass of the object, m = 6.55 kg

Let, the Force by bicep be, $F_b$

under the motionless condition, the net moment about elbow is zero

thus,

$F_b$ × 0.0215 - Mg × (0.465/2) - mg × (0.465) = 0

or

$F_b$ × 0.0215 - 2.45 × 9.8 × (0.465/2) - 6.55 × 9.8 × (0.465) = 0

or

$F_b$ × 0.0215 - 5.582 - 29.848 = 0

or

$F_b$  = 1647.92 N

hence, the force exerted by the elbow is 1647.92 N

Answer 2

This force exerted by the biceps muscles has been calculated as;

F_b = 1648 N

The image showing the bones of the forearm and biceps is missing and I have attached it.

We are given;

Length of forearm; L = 0.465 m

Mass of forearm; m_f = 2.45 kg

Mass of the object; m_o = 6.55 kg

Distance; d_f = 2.15 cm = 0.0215 m

• Under the equilibrium condition, we can calculate the  force exerted by the biceps muscle by;

Taking moment about the elbow (E) and equating to zero, we have;

(F_b × d_f) - (m_f × g × L/2) - (m_o × g × L) = 0

Where F_b is the force exerted by the biceps muscles.

• Plugging in the relevant values, we have;

0.0215F_b - (2.45 × 9.8 × (0.465/2)) - (6.55 × 9.8 × 0.465) = 0

Rearranging gives;

0.0215F_b = (2.45 × 9.8 × (0.465/2)) + (6.55 × 9.8 × 0.465)

Multiplying out the bracket gives us;

0.0215F_b = 35.430675

F_b = 35.430675/0.0215

F_b ≈ 1648 N

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