Question

When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.40 kg .If the biceps is connected to the forearm a distance dbiceps = 3.50 cm from the elbow, how much force Fbiceps must the biceps exert to hold a 850 g ball at the end of the forearm at distance dball = 33.0 cmfrom the elbow, with the forearm parallel to the floor
Part A
Although it is uncommon, orthopedic surgeons have seen patients who have torn their biceps tendon when forces of greater than 390 N have been exerted on the tendon with the arm bent at the elbow so that the forearm is parallel to the ground. What is the minimum mass of a ball Mball such a patient might have held to cause the biceps tendon to tear?

Answer 1

1 Part:  According to the given data provided in the question, the center of masses of forearm = 1.40 kg and ball = 0.850 kg is;

$C.M.=\frac{(1.40)(0.035)+(0.850)(0.33)}{1.40+0.850}=0.146\: m$

Now the force exerted by biceps can be calculated as;

  $\tau =(1.40)(9.81)(0.035)+(0.850)(9.81)(0.33)$

$\Rightarrow F_{biceps}(C.M.) =3.23$

$\Rightarrow F_{biceps}=\frac{3.23}{0.146}=22.12\: N$

2 Part:  According to the question, a patient will torn their biceps tendon when forces is greater than 390 N. Therefore the minimum mass of a ball M_ball such a patient might have held to cause the biceps tendon to tear is;

$F_{biceps}\left (\frac{(1.40)(0.035)+(M_{ball})(0.33)}{1.40+M_{ball}} \right ) =(1.40)(9.81)(0.035)+(M_{ball})(9.81)(0.33)$

  $\Rightarrow 390\left (\frac{0.049+(M_{ball})(0.33)}{1.40+M_{ball}} \right ) =0.481+(M_{ball})3.237$

  $\Rightarrow 19.11+128.7(M_{ball})=0.67+4.53(M_{ball})+0.481(M_{ball})+3.237(M_{ball})^{2}$

  $\Rightarrow 3.237(M_{ball})^{2}-123.689(M_{ball})-18.44=0$

  $\Rightarrow (M_{ball})^{2}-38.21(M_{ball})-5.7=0$

  $\Rightarrow M_{ball}=\frac{38.21\pm \sqrt{(38.21)^{2}-4(1)(-18.44)}}{2}$

  $\Rightarrow M_{ball}=\frac{38.21\pm 39.16}{2}=38.685\: or\: -0.475$

Hence the mass of the ball is 38.685 kg.