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 2.15 cm beyond the joint. Assume the forearm has a mass of 2.25 kg and a length of 0.425 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
1 answer:
Answer:
The force exerted by the biceps is 143.8 kgf.
Explanation:
To calculate the force exerted by the biceps, we calculate the momentum in the elbow.
This momentum has to be zero so that her forearm remains motionless.
Being:
W: mass weight (6.15 kg)
d_W= distance to the mass weight (0.425 m)
A: weight of the forearm (2.25 kg)
d_A: distance to the center of mass of the forearm (0.425/2=0.2125 m)
H: force exerted by the biceps
d_H: distance to the point of connection of the biceps (0.0215 m)
The momemtum is:
The force exerted by the biceps is 143.8 kgf.
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Answer:
h = 3.5 m
Explanation:
First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 3.5 m
vf = final speed = ?
vi = initial speed = 0 m/s
Therefore,
Now, we will apply the law of conservation of momentum:
where,
m₁ = mass of colliding ball = 3.6 kg
m₂ = mass of ball on the other end = 3.6 kg
v₁ = vf = final velocity of ball while collision = 8.3 m/s
v₂ = vi = initial velocity of other end ball = ?
Therefore,
Now, we again use the third equation of motion for the upward motion of the ball:
where,
g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)
h = height = ?
vf = final speed = 0 m/s
vi = initial speed = 8.3 m/s
Therefore,
<u>h = 3.5 m</u>