Torque in bicep, please help (Urgent)

  • Thread starter laddoo12
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  • #1
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Homework Statement



a man holds a 5.00kg weight in his hand. His arm is represented here by a pair of thin rods attached with a hinge, with a cord for the bicep muscle. The forearm makes a 20 degree angle with the horizontal, and the bicep (which is attached 4.0 cm away from the elbow joint) makes a 55 degree angle from the horizontal. assumme that the forearm has a mass of 1.2 Kg and is 36cm long, and has its centre of mass midway through its length. find the tension in the bicep muscle and the components of force in the elbow joint!

DIAGRAM:
http://farm9.staticflickr.com/8445/7797836656_015d170c93.jpg

Homework Equations


Torque =F*d


The Attempt at a Solution



Okay so ill show you two ways i've don't it because I'm confused about how its actually done

1st way it think it may be wrong

Sin(55)(4)Ft=(sin70*9.8*1.2)(18)+(sin70*9.8*5*36)

solveing for Ft = 556N

Attempt two

sin(55+20)*4(ft)=(sin70*9.8*1.2)(18)+(sin70*9.8*5*36)

Ft= 480N

and the vertical component =

9.8*1.2+5*9.8-480*sin55=-332N or 332N to the vertical

horizontal component = sin55*480=393N

Am i going wrong anywhere, which method is correct PLEASE HELP!
 

Answers and Replies

  • #2
SammyS
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Homework Statement



a man holds a 5.00kg weight in his hand. His arm is represented here by a pair of thin rods attached with a hinge, with a cord for the bicep muscle. The forearm makes a 20 degree angle with the horizontal, and the bicep (which is attached 4.0 cm away from the elbow joint) makes a 55 degree angle from the horizontal. assumme that the forearm has a mass of 1.2 Kg and is 36cm long, and has its centre of mass midway through its length. find the tension in the bicep muscle and the components of force in the elbow joint!

DIAGRAM:
http://farm9.staticflickr.com/8445/7797836656_015d170c93.jpg

Homework Equations


Torque =F*d

The Attempt at a Solution



Okay so I'll show you two ways I've done it because I'm confused about how its actually done

1st way it think it may be wrong

Sin(55)(4)Ft=(sin70*9.8*1.2)(18)+(sin70*9.8*5*36)

solveing for Ft = 556N

Attempt two

sin(55+20)*4(ft)=(sin70*9.8*1.2)(18)+(sin70*9.8*5*36)

Ft= 480N

and the vertical component =

9.8*1.2+5*9.8-480*sin55=-332N or 332N to the vertical

horizontal component = sin55*480=393N

Am i going wrong anywhere, which method is correct PLEASE HELP!
In your first way, you are incorrectly calculating the torque that the bicep produces about the elbow joint.

The second way looks fine for torque.

Here's the graphic for anyone viewing this:
7797836656_015d170c93.jpg
 
  • #3
ehild
Homework Helper
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In your first way, you are incorrectly calculating the torque that the bicep produces about the elbow joint.

The second way looks fine for torque.

Here's the graphic for anyone viewing this:
7797836656_015d170c93.jpg
The torque of the biceps on the forearm about the elbow does not depend on the position of the forearm. The torque is produced by the component of force which is perpendicular to the arm, and that is Ftsin(55°), the torque is τ=4Ftsin(55).
The first way is correct.

ehild
 
  • #4
SammyS
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The torque of the biceps on the forearm about the elbow does not depend on the position of the forearm. The torque is produced by the component of force which is perpendicular to the arm, and that is Ftsin(55°), the torque is τ=4Ftsin(55).
The first way is correct.

ehild
The bicep makes an angle of (55° + 20°) to fore arm, so the component of the force perpendicular to the fore arm is Ttsin(75°) .

SammyS
 
  • #5
ehild
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Is that 55°angle measured from the forearm as in the figure or from the horizontal as written in the text?

ehild
 

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