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4 years ago

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The bones of the forearm (radius and ulna) are hinged to the humerus at the elbow. The biceps muscle connects to the bones of th

e 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?

2 answers:

julsineya [31]4 years ago

5 0

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

SVETLANKA909090 [29]4 years ago

4 0

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

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