- #1
faoltaem
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would someone be able to tell me if this is right or what I've done wrong? thankyou
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?
[tex]\tau[/tex] = Fx = mgx
[tex]\tau[/tex] = Fsin[tex]\theta[/tex]
[tex]\Sigma[/tex][tex]\tau[/tex] = 0
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
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