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:

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

the net moment about elbow is zero

Moment = force *distance

thus,

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

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

=

× 0.0215 - 5.582 - 29.848 = 0

or

 = 1647.92 N

hence, the force exerted by the elbow is 1647.92 N

Answer 2

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