Q68P - University Physics with Modern Physics

Question

In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm $\mathbf{3}\mathbf{.}\mathbf{80}\mathit{c}\mathit{m}$  from the elbow joint. Assume that the person’s hand and forearm together weigh $\mathbf{15}\mathbf{.}\mathbf{0}\mathit{N}$  and that their center of gravity is $\mathbf{15}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$  from the elbow (not quite halfway to the hand). The forearm is held horizontally at a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the forearm. (a) Draw a free-body diagram for the forearm, and find the force exerted by the biceps when the hand is empty. (b) Now the person holds an $\mathbf{80}\mathbf{.}\mathbf{0}\mathit{N}$  weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is $\mathbf{33}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$   from the elbow. Draw a free-body diagram for the forearm, and find the force now exerted by the biceps. Explain why the biceps muscle needs to be very strong. (c) Under the conditions of part (b), find the magnitude and direction of the force that the elbow joint exerts on the forearm. (d) While holding the  $\mathbf{80}\mathbf{.}\mathbf{0}\mathit{N}$ weight, the person raises his forearm until it is at an angle of  $\mathbf{53}\mathbf{.0}\mathbf{\text{°}}$ above the horizontal. If the biceps muscle continues to exert its force perpendicular to the forearm, what is this force now? Has the force increased or decreased from its value in part (b)? Explain why this is so, and test your answer by doing this with your own arm.

Step-by-Step Solution

Answer

1. The forearm diagram is given by,

 

 

The force exerted by the biceps when the hand is empty is 59.21 N .

1.  The forearm diagram is given by,

The force now exerted by the biceps is  753.95 N 

1.  ${\mathrm{O}}_{\mathrm{x}}=0\mathrm{N}$ and ${\mathrm{O}}_{\mathrm{y}}=658.95 \mathrm{N}$ direction of ${\mathrm{O}}_{\mathrm{y}}$ is downwards.
2.  The force ${\mathrm{F}}_{\mathrm{B}}=453.74\mathrm{N}$.

       The value of ${\mathrm{F}}_{\mathrm{B}}$ is decreased from the value found in part (b) because the arm levers of   and the                   force 80.0 N are reduced by the factor $\cos \theta$ .

Step 1: Second condition of equilibrium

Which is says that, For the body to continue to be nonrotating, the net external torque around any point on the body must be zero.

$\mathbf{\sum }_{\mathbf{ }}\overrightarrow{\mathbf{\tau }}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\left(}\mathbf{1}\mathbf{\right)}$                                   (1)

Where $\overrightarrow{\mathbf{\tau }}$ is net external torque.

Step 2: Identification of given data

Here we have given that the biceps muscle is attached to the forearm   from the elbow joint. So, .

The person’s hand and forearm together weigh  . So, a = 0.038 m 

Their center of gravity is 15.0 cm from the elbow. So, b = 0.15 m   

Step 3: Drawing a free-body diagram for the forearm, and find the force exerted by the biceps when the hand is empty.

(a)

The free-body diagram for the forearm is given by,

${\mathrm{F}}_{\mathrm{B}}$  is the force exerted by the biceps on the forearm.

${\mathrm{O}}_{\mathrm{x}}$  is the horizontal force exerted by the elbow.

${\mathrm{O}}_{\mathrm{y}}$  is the vertical force exerted by the elbow

Now, from equation (1)

$\begin{array}{r}\sum {\tau }_0=0\\ \left(0\right){\mathrm{O}}_{\mathrm{x}}+\left(0\right){\mathrm{O}}_{\mathrm{y}}+{\mathrm{aF}}_{\mathrm{B}}-\mathrm{bW}=0\end{array}$

             $\begin{array}{r}{\mathrm{F}}_{\mathrm{B}}=\frac{\mathrm{bW}}{\mathrm{a}}\\ =\frac{\left(0.150\mathrm{m}\right)\left(15.0\mathrm{N}\right)}{0.0380\mathrm{m}}\\ =59.21\mathrm{N}\end{array}$

So, the force exerted by the biceps when the hand is empty is 59.21 N .

Step 4: Draw a free-body diagram for the forearm, and find the force now exerted by the biceps. Explain why the biceps muscle needs to be very strong.

(b)

Here holding the 80.0 N weight.

So, the Free-body diagram for the forearm is given by,

${\mathrm{F}}_{\mathrm{B}}$  is the force exerted by the biceps on the forearm.

${\mathrm{O}}_{\mathrm{x}}$  is the horizontal force exerted by the elbow.

${\mathrm{O}}_{\mathrm{y}}$  is the vertical force exerted by the elbow

Now, from equation (1)

$\begin{array}{r}\sum {\tau }_0=0\\ \left(0\right)O_x+\left(0\right)O_y+a{F}_B-bW-c\left(80.0\mathrm{N}\right)=0\\ F_B=\frac{bW+c\left(80.0\mathrm{N}\right)}{a}\\ F_B=\frac{\left(0.150\mathrm{m}\right)\left(15.0\mathrm{N}\right)+\left(0.33\mathrm{m}\right)\left(80.00\mathrm{N}\right)}{0.0380\mathrm{m}}\\ {\mathrm{F}}_{\mathrm{B}}=753.95\mathrm{N}\end{array}$

The force now exerted by the biceps is $753.95 \mathrm{N}$

In order to balance the torques about the elbow of   and the force of 80.0 N , ${\mathrm{F}}_{\mathrm{B}}$ must be large because its arm level about the elbow is short.

Step 5: find the magnitude and direction of the force that the elbow joint exerts on the forearm.

(c)

Now, by equation (1) and a free-body diagram for the forearm in step 5,

$\begin{array}{r}\sum F_x=0\\ -O_x=0\\ \sum F_y=0\\ -O_y+F_B-W-80.0\mathrm{N}=0\\ O_y=753.95\mathrm{N}-15.0\mathrm{N}-80.0\mathrm{N}\\ =658.95\mathrm{N}\end{array}$

the direction of ${\mathrm{O}}_{\mathrm{y}}$ is downwards.

Step 6: Finding the force when the person raises his forearm until it is at an angle of above the horizontal. Check that force is increased or decreased from its value in part (b).

(d)

Here While holding the 80.0 N weight, the person raises his forearm until it is at an angle of $53.0°$ above the horizontal.

So, its free body diagram for the forearm is given by,

Now, by equation (1)

$\sum _{ }{\tau }_0=0$

$\begin{array}{r}\left(0\right)O_x+\left(0\right)O_y+a{F}_B-bW-c(\cos \theta )\left(80.00N\right)=0\\ F_B=\frac{bW+c\left(80.00N\right)}{a}\cos \theta \\ F_B=\frac{\left(0.150\mathrm{m}\right)\left(15.0\mathrm{N}\right)+\left(0.33\mathrm{m}\right)\left(80.00\mathrm{N}\right)}{0.0380\mathrm{m}}\cos {53}^{\circ }\\ F_B=453.74\mathrm{N}\end{array}$

So, the value of ${\mathrm{F}}_{\mathrm{B}}$ is decreased from the value found in part (b) because the arm levers of W and the force 80.0 N are reduced by the factor $\mathrm{cos\theta }$.