When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.30 kg . If the biceps is connected to the forearm a distance dbiceps = 3.00 cm from the elbow, how much force Fbiceps must the biceps exert to hold a 850 g ball at the end of the forearm at distance dball = 34.0 cm from the elbow, with the forearm parallel to the floor? How much force Felbow must the elbow exert?

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Answer:

The answers to the questions are;

The force the biceps must exert to hold a 850 g ball at the end of the forearm at distance dball = 34.0 cm from the elbow, with the forearm parallel to the floor is 166.77 N Upwards.

The force the elbow must  exert is 145.6785 N downwards.

Explanation:

We are required to analyze the system using the principle of moments as follows.

Mass of forearm = 1.30 kg

Weight of forearm = mass × acceleration due to gravity = 1.30 kg × 9.81  m/s²= 12.753 N

Location of point of action of weight of forearm = midpoint  17 cm

Forearm to biceps distance = 3.00 cm from the elbow.

Mass of the ball = 850 g

Weight of ball = mass × acceleration due  to gravity = 0.850 kg × 9.81 = 8.3385 N

Position of the ball = end of the forearm

Therefore, taking moment about the elbow, we have

∑Mₓ = 0, we take clockwise moment to be positive, therefore

8.3385 N× 34.0 cm + 12.753 N× 17 cm - [tex]F_{biceps}[/tex]×3.00 cm = 0

Therefore [tex]F_{biceps}[/tex]×3.00 cm = 500.31 N cm

Therefore [tex]F_{biceps}[/tex] = (500.31 N·cm)/(3.00 cm) = 166.77 N Upwards

The force exerted by the elbow is given by,

Taking moment about the point of contact between the biceps and the forearm, we get

[tex]F_{elbow}[/tex] × 3.00 cm = 8.3385 N× 31.0 cm + 12.753 N× 14 cm

[tex]F_{elbow}[/tex] = 145.6785 N downwards.

The biceps force (Fbiceps) and the elbow force(Felbow) is mathematically given as

  • Fbiceps= 166.77 N towards the upward direction
  • Felbow= 145.6785 N  towards the downwards direction

What are the biceps force and the elbow force?

Generally, the equation for the Weight of the forearm  is mathematically given as

W=mass ×  gravity

W = 1.30 kg × 9.81  m/s^2

W= 12.753 N

The moment about the elbow is

Since the sum of moments about the elbow is equaled 0

8.3385 N× 34.0 cm + 12.753 N× 17 cm -Fbiceps ×3.00 cm = 0

Fbiceps ×3.00 cm = 500.31 N cm

Fbiceps = (500.31 N·cm)/(3.00 cm)

Fbiceps= 166.77 N towards the upward direction

the sum of moments about the point b/w biceps and forearm is

Felbow × 3.00 cm = 8.3385 N× 31.0 cm + 12.753 N× 14 cm

Felbow= 145.6785 N  towards the downwards direction

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