Question

biophysics 1 problem set 4

1. two fishermen are exerting a force of 200 n on a paddle while rowing in the middle of a lake. if the hands of the fishermen are located 2 m and 3 m from the pivot point of the paddle, then how much is the torque produced?
2. as part of his daily workout, a person lifts a 10 - kg dumbbell in each hand. each of his forearms and hands has a combined mass of 4 kg and a total length of 0.5 m. the biceps muscle acts vertically and is attached to the forearm at a point 5.50 cm from the elbow joint. assuming that the center of gravity of the forearm - hand system is at its midpoint, determine the force exerted by the biceps muscle needed to hold the forearm horizontally.

Explanation:

Step1: Calculate the torque due to each fisherman

Torque formula is $\tau = rF\sin\theta$. Assuming the force is perpendicular to the paddle ($\theta = 90^{\circ}$, $\sin\theta=1$).
For the first fisherman: $\tau_1=r_1F_1$, where $r_1 = 2m$ and $F_1=200N$. So $\tau_1=2\times200 = 400N\cdot m$.
For the second fisherman: $\tau_2=r_2F_2$, where $r_2 = 3m$ and $F_2=200N$. So $\tau_2=3\times200=600N\cdot m$.

Step2: Calculate the total torque

Since the torques are in the same rotational direction (assuming they are rowing in the same rotational sense), the total torque $\tau=\tau_1+\tau_2$.
$\tau=400 + 600=1000N\cdot m$.

Step3: Analyze the forearm - dumbbell system for the second problem (using torque equilibrium)

Let the force exerted by the biceps be $F$.
The weight of the dumbbell $W_d=mg$, where $m = 10kg$ and $g = 9.8m/s^2$. So $W_d=10\times9.8 = 98N$.
The weight of the forearm - hand system $W_f=mg$, where $m = 4kg$. So $W_f=4\times9.8=39.2N$.
The torque due to the dumbbell: $\tau_d=r_dW_d$, $r_d = 0.5m$.
The torque due to the forearm - hand system: $\tau_f=r_fW_f$, $r_f=\frac{0.5}{2}=0.25m$.
The torque due to the biceps: $\tau_b=r_bF$, $r_b = 0.055m$.
In torque equilibrium $\sum\tau = 0$ (counter - clockwise torques = clockwise torques). Assuming the biceps torque is counter - clockwise and the other torques are clockwise.
$r_bF=r_dW_d+r_fW_f$.
Substitute the values: $0.055F=0.5\times98+0.25\times39.2$.
First calculate the right - hand side: $0.5\times98+0.25\times39.2=49 + 9.8=58.8$.
Then solve for $F$: $F=\frac{58.8}{0.055}\approx1069.09N$.

Answer:

1. The torque produced by the two fishermen is $1000N\cdot m$.
2. The force exerted by the biceps muscle is approximately $1069.09N$.