Question

practice it
use the worked example above to help you solve this problem. a ( w = 47.7 mathrm{~n}(10.7 mathrm{lb}) ) weight is held in a
persons hand with the forearm horizontal, as shown in the figure. the biceps muscle is attached ( d=)
0.02962 ( mathrm{m} ) from the joint, and the weight is ( l=0.345 mathrm{~m} ) from the joint. find the upward force ( overrightarrow{mathrm{f}} ) exerted
by the biceps on the forearm (the ulna) and the downward force ( overrightarrow{mathrm{r}} ) exerted by the humerus on the
forearm, acting at the joint. neglect the weight of the forearm.
( f=quad mathrm{n} )
( r=quad mathrm{n} )
exercise
hints: getting started | im stuck!
use the values from practice it to help you work this exercise. suppose you wanted to limit the force
acting on your joint to a maximum value of ( 8.14 \times 10^{2} mathrm{~n} ).
(a) under these circumstances, what maximum weight would you attempt to lift?
( mathrm{n} )
(b) what force would your biceps apply while lifting this weight?
( mathrm{n} )

Explanation:

Step1: Analyze forces and torques for Practice It

• Torque equilibrium:
• The torque due to the biceps force \(F\) about the joint is \(\tau_F = Fd\) (counter - clockwise, positive). The torque due to the weight \(W\) is \(\tau_W=-Wl\) (clockwise, negative). In equilibrium, \(\sum\tau = 0\), so \(Fd−Wl = 0\).
• Solving for \(F\): \(F=\frac{Wl}{d}\).
• Substitute \(W = 47.7\space N\), \(l = 0.345\space m\), and \(d=0.02962\space m\) into the formula: \(F=\frac{47.7\times0.345}{0.02962}\).
• \(F=\frac{16.4565}{0.02962}\approx556\space N\).
• Force equilibrium:
• In the vertical direction, \(\sum F_y = 0\). So \(F - W- R=0\), and \(R=F - W\).
• Substitute \(F = 556\space N\) and \(W = 47.7\space N\) into the formula: \(R=556 - 47.7=508.3\space N\).

Step2: Analyze forces and torques for Exercise (a)

• Torque and force equations:
• From torque equilibrium \(Fd−Wl = 0\), so \(F=\frac{Wl}{d}\). From force equilibrium \(F - W- R = 0\), so \(F=W + R\).
• Substitute \(F=\frac{Wl}{d}\) into \(F=W + R\): \(\frac{Wl}{d}=W + R\).
• Rearrange for \(W\): \(\frac{Wl}{d}-W=R\), \(W(\frac{l}{d}-1)=R\).
• Given \(R = 8.14\times10^{2}\space N\), \(l = 0.345\space m\), \(d = 0.02962\space m\).
• \(\frac{l}{d}-1=\frac{0.345}{0.02962}-1\approx11.65 - 1=10.65\).
• Then \(W=\frac{R}{\frac{l}{d}-1}=\frac{814}{10.65}\approx76.4\space N\).

Step3: Analyze forces and torques for Exercise (b)

• Use torque formula:
• From \(Fd−Wl = 0\), \(F=\frac{Wl}{d}\).
• Substitute \(W = 76.4\space N\), \(l = 0.345\space m\), \(d = 0.02962\space m\).
• \(F=\frac{76.4\times0.345}{0.02962}=\frac{26.358}{0.02962}\approx890\space N\).

Answer:

Practice It:
\(F = 556\space N\)
\(R = 508\space N\)
Exercise:
(a) \(76.4\space N\)
(b) \(890\space N\)