How to calculate torque for bicep in rotational motion?

In summary, the bicep muscle needs to apply a torque of 2Nm to hold the forearm of mass 1.10 kg and length 30 cm horizontally. The bicep needs to apply a torque of 60.47865Nm to hold a 20 kg weight in one hand at an angle of 20\circ below horizontal.
  • #1
faoltaem
31
0
would someone be able to tell me if this is right or what I've done wrong? thankyou

Homework Statement



a) What torque does the bicep muscle have to apply to hold the forearm of mass 1.10 kg and length 30 cm horizontally? Assume the arm is a rod of uniform mass density.
b) What torque does the bicep need to apply if the forearm is to hold a 20 kg weight in one hand at an angle of 20[tex]\circ[/tex] below horizontal?

Homework Equations



[tex]\tau[/tex] = Fx = mgx
[tex]\tau[/tex] = Fsin[tex]\theta[/tex]
[tex]\Sigma[/tex][tex]\tau[/tex] = 0

The Attempt at a Solution



a) m = 1.1kg x = 15cm = 0.15m
[tex]\tau[/tex] = Fx = mgx = 1.1 x 9.81 x 0.15
= 1.61865
= 2Nm

b) m[tex]_{o}[/tex] = 20kg [tex]\theta[/tex] = 20[tex]\circ[/tex]
[tex]\Sigma[/tex][tex]\tau[/tex] = 0
[tex]\tau[/tex][tex]_{1}[/tex] + [tex]\tau[/tex][tex]_{2}[/tex] = 0
[tex]\tau[/tex][tex]_{a}[/tex] + [tex]\tau[/tex][tex]_{o}[/tex] = 0 a=arm o=object
[tex]\tau[/tex][tex]_{o}[/tex] = -[tex]\tau[/tex][tex]_{a}[/tex]
Fsin[tex]\theta[/tex]r = -mgx
F x sin20 x 0.3 = -(1.1 x –9.81 x 0.15)
0.1026F = 1.61865

F = [tex]\frac{1.61865}{0.1026}[/tex]
= 15.775
= 16Nm
 
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  • #2
Part a is correct. One uses the CM of the arm to determine the moment about the elbow joint.

The logic on part b is not correct. The moment of the object and the moment of the arm work together against the bicep. The torque provided by the bicep must equal the sum of torques of the arm and object.
 
  • #3
sorry i forgot about the 3rd force
[tex]\Sigma[/tex][tex]\tau[/tex] = 0
[tex]\tau[/tex]1 + [tex]\tau[/tex]2 + [tex]\tau[/tex]3 = 0
[tex]\tau[/tex]1 + [tex]\tau[/tex]2 = -[tex]\tau[/tex]3

but I'm a little unsure about the force of [tex]\tau[/tex]2 (which i have as the object)

is it: -[tex]\tau[/tex]3 = [tex]\tau[/tex]1 + [tex]\tau[/tex]2
= m1gx1 + m2gx2
= 1.1 x 9.81 x 0.15 + 20 x 9.81 0.3
= 60.47865N
[tex]\tau[/tex]3 = -60.47865N
= -6 x 10[tex]^{2}[/tex]N
 
  • #4
Be careful with units, torque is N-m, as opposed to force which uses N.

Don't forget the angle 20°, which influences the force normal to the moment arm.
 
  • #5
ok so
[tex]\tau[/tex]1 = Fsin[tex]\theta[/tex]
= mgsin[tex]\theta[/tex]
= 1.1 x -9.81 x sin 20 x 0.15
= -0.5536Nm

[tex]\tau[/tex]2 = Fsin[tex]\theta[/tex]
= mgsin[tex]\theta[/tex]
= 20 x -9.81 x sin 20 x 0.3
= -20.131Nm

-[tex]\tau[/tex]3 = [tex]\tau[/tex]1 + [tex]\tau[/tex]2
= -0.5536 + -20.131
= -20.685Nm

therefore [tex]\tau[/tex]3 = 20.685Nm
or 20.685Nm in an anticlockwise direction
 
  • #6
I believe one wants the cosine of the angle (between the weight and the normal to the moment arm) in this case or the sine of the angle between the force and the moment arm (which would be 90° + 20°).
 
  • #7
so if i change sin 20 to cos 20 (or sin 110) i get:

[tex]\tau[/tex]1 = Fsin
= mgsin
= 1.1 x -9.81 x cos 20 x 0.15
= -1.521Nm

[tex]\tau[/tex]2 = Fsin
= mgsin
= 20 x -9.81 x cos 20 x 0.3
= -55.310Nm

-3 = 1 + 2
= -1.521 + -55.310
= -56.831Nm

therefore 3 = 56.831Nm
or 56.831Nm in an anticlockwise direction

is this the correct answer?
(thankyou for all your help)
 
  • #8

1. What is rotational motion of the bicep?

The rotational motion of the bicep refers to the movement of the bicep muscle around its axis of rotation, which is the elbow joint. This motion allows for the bending and straightening of the arm.

2. How does rotational motion of the bicep differ from other types of motion?

Rotational motion of the bicep is a type of angular motion, which involves movement around an axis rather than in a straight line. This is different from linear motion, which involves movement along a straight path.

3. What are the factors that influence the rotational motion of the bicep?

The rotational motion of the bicep is influenced by the contraction of the bicep muscle, the strength and coordination of surrounding muscles, and the resistance or weight being lifted.

4. How does rotational motion of the bicep contribute to everyday activities?

The rotational motion of the bicep is essential for everyday activities such as lifting objects, reaching for items, and performing daily tasks that require the use of the arm and hand.

5. What are some common injuries associated with rotational motion of the bicep?

Some common injuries associated with rotational motion of the bicep include strains, tears, and dislocations of the bicep muscle or tendon. These injuries can occur from overuse, sudden movements, or improper lifting techniques.

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