Q27P In Fig. 12-44, a 15 kg block is ... [FREE SOLUTION]

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Chapter 12: Q27P (page 347)

In Fig. 12-44, a 15 kg block is held in place via a pulley system. The person鈥檚 upper arm is vertical; the forearm is at angle $胃 = 30^{o}$ with the horizontal. Forearm and hand together have a mass of 2.0 kg, with a center of mass at distance $d_{1} = 15 鈥塩尘$ from the contact point of the forearm bone and the upper-arm bone(humerus). The triceps muscle pulls vertically upward on the forearm at distance $d_{2} = 2 .5 鈥塩尘$ behind that contact point. Distance $d_{3}$ is 35 cm. What are the (a)magnitude and (b) direction (up or down) of the force on the forearm from the triceps muscle and the (c) magnitude and (d) direction (up or down) of the force on the forearm from the humerus?

Short Answer

Expert verified

a)The magnitude of the force on the forearm from the triceps muscles is $F_{t r i c e p s} = 1.94 脳 10^{3} 鈥塏$ , .

b) The direction of the force on the forearm from the triceps muscles isupward.

c) The magnitude of the force on the forearm from the humerus is, $F_{h u m e r u s} = 2.1 脳 10^{3} 鈥塏$ .

d) The direction of the force on the forearm from the humerus is downward.

Step by step solution

Understanding the given information

i) Mass of block, M= 15 kg.

ii) The forearm makes an angle with the horizontal, $胃 = 30^{鈭�}$

iii) Mass of forearm and hand together, m = 2kg.

iv) Distance of center of mass of forearm and hand together from the contact point of forearm bone and upper bone, $d_{1} = 15 鈥塩尘$ .

v) Distance of force by triceps muscles from the contact point of forearm bone and upper bone, $d_{2} = 2.5 鈥塩尘$ .

vi) The distance between fingers holds the rope and the contact point, $d_{3} = 35 鈥塩尘$ .

Concept and formula used in the given question

You can use the concept of static equilibrium and torque to find the force from the triceps on the forearm. Also, you can use the net force along the x and y directions for equilibrium to find the forces of the triceps and humerus.From the resulting answer, you can determine the direction of the forces.

(a) Calculation for the force

The given forces are all vertical, and the distances are measured along the x-axis at theangle $胃 = 30^{鈭�}$ .

Here, if we use trigonometry, the trigonometric factors cancel out. So, we can write the torque about the contact point and solve it for the force.

The net torque is calculated as,

$鈭� F_{t r i c e p s} d_{2} 鈭� 2 (9.8) d_{1} + 15 (9.8) d_{2} = 0$

Rearranging for $F_{t r i c e p s}$ we get,

$F_{t r i c e p s} = \frac{15 (9.8) d_{2} 鈭� 2 (9.8) d_{1}}{d_{2}}$

$\begin{matrix} F_{t r i c e p s} = \frac{15 鈥尘脳 (9.8 {鈥尘/s}^{2}) (0.35 鈥塳驳) 鈭� 2 鈥尘脳 (9.8 {鈥尘/s}^{2}) (0.15 鈥塳驳)}{0.025 鈥尘} \\ = 1.94 脳 10^{3} 鈰� (1 鈥塳驳鈰� {m/s}^{2} 脳 \frac{1 鈥塏}{1 鈥塳驳鈰� {m/s}^{2}}) \\ = 1.94 脳 10^{3} 鈥塏 \end{matrix}$

Thus, the magnitude of the force on the forearm from the triceps muscles is,

F_{t r i c e p s} = 1.94 脳 10^{3} 鈥塏

(b) Direction of the force on the triceps muscle

From part (a), we got a positive value. So, the direction is upwards.

Thus, the force on the triceps muscle will be in the upward direction.

(c) Calculation for the force

For equilibrium, the vertical forces are balanced. We can thus write the equation as:

$\begin{matrix} F_{t r i c e p s} + F_{h u m e r u s} + 15 (9.8) 鈭� 2 (9.8) = 0 \\ 1.94 脳 10^{3} + F_{h u m e r u s} = 19.6 鈭� 147 \\ F_{h u m e r u s} = 鈭� 127.4 鈭� 1.94 脳 10^{3} \\ = 鈭� 2067.4 鈥塏 \\ = 鈭� 2.1 脳 10^{3} 鈥塏 \end{matrix}$