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 2.15 cm beyond the joint. Assume the forearm has a mass of 2.35 kg and a length of 0.445 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 6.15 kg so that her forearm remains motionless, what is the force exerted by the biceps muscle

Answer 1

Answer:

The force exerted by the biceps muscle is 1485.58 N.

Explanation:

Given that,

Length = 0.445 m

Mass of forearm = 2.15 cm

Mass of the object = 6.15

Distance = 2.15 cm

We need to calculate the  force exerted by the biceps muscle

Under the equilibrium condition

Moment about E = 0

Using free body diagram of muscle

$F_{b}\times d_{f}-mg\times\dfrac{l}{2}-mg\times l=0$

Where, $F_{b}$ = force by biceps

put the value into the formula

$F_{b}\times0.0215-mg\times\dfrac{0.445}{2}-mg\times0.445=0$

$F_{b}\times0.0215=2.35\times9.8\times0.2225+6.15\times9.8\times0.445$

$F_{b}\times 0.0215=31.94$

$F_{b}=\dfrac{31.94}{0.0215}$

$F_{b}=1485.58\ N$

Hence, The force exerted by the biceps muscle is 1485.58 N.

Answer 2

Answer:

151.6 kg-weight

Explanation:

Torque = force x perpendicular distance of direction of force

We shall take torque of forces about the elbow joint. If T be tension in the forearm muscle , Torque about elbow is

$T\times .0215$

Torque of weight of bone about elbow

$=2.35\times \frac{0.445}{2}$

=.5228 Nm

Torque of weight held about the elbow

$=6.15\times 0.445\\=2.7367$

For rotational equilibrium

$T\times .0215$ =.5228 Nm+2.7367

T = 151.60 Kg-wt

=