In Fig. 12-44, a 15 kg block is held in place via a pulley...

In Fig. $12-44,$ a 15 kg block is held in place via a pulley system. The person's upper arm is vertical; the forearm is at angle $\theta=30^{\circ}$ with the horizontal. Forearm and hand together have a mass of 2.0 $\mathrm{kg}$ , with a center of mass at distance $d_{1}=15 \mathrm{cm}$ 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 $d_{2}=2.5 \mathrm{cm}$ behind that contact point. Distance $d_{3}$ is 35 $\mathrm{cm} .$ What are the (a) magnitude and (b) direction (up or down) of the force on the forearm from the tri- ceps muscle and the (c) magnitude and (d) direction (up or down) of the force on the forearm from the humerus?

In Fig. $12-44,$ a 15 kg block is held in place via a pulley system. The
person's upper arm is vertical; the forearm is at angle $\theta=30^{\circ}$ with the horizontal. Forearm and hand together have a
mass of 2.0 $\mathrm{kg}$ , with a center of mass at distance $d_{1}=15 \mathrm{cm}$ 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 $d_{2}=2.5 \mathrm{cm}$ behind that
contact point. Distance $d_{3}$ is 35 $\mathrm{cm} .$ What are the (a) magnitude and
(b) direction (up or down) of the force on the forearm from the tri-
ceps muscle and the (c) magnitude and (d) direction (up or down) of
the force on the forearm from the humerus?

Key Concepts

Biomechanics of Force Application

In biomechanics, especially in musculoskeletal systems, the analysis of how muscles apply forces about joints involves understanding leverage, moment arms, and the direction of forces. This allows for the examination of mechanical advantages and the resulting stresses on bones and muscles during movement or static postures.

Free-Body Diagram Analysis

Free-body diagrams are graphical representations used to isolate a body and illustrate all the external forces and moments acting on it. This systematic approach is essential for setting up the equilibrium equations necessary to solve for unknown forces and torques.

Lever Arm Concept

The lever arm is the perpendicular distance from the axis of rotation to the line of action of a force. It is a key factor in torque calculations because even a small force can produce a large torque if applied far from the pivot, affecting the balance of the system.

Torque (Moment) Calculation

Torque is the rotational equivalent of force and is calculated as the product of a force and its lever arm (perpendicular distance from the pivot point). It plays a crucial role in determining how forces produce rotation in systems in equilibrium.

Static Equilibrium

This concept deals with objects at rest where the sum of all forces and torques (moments) acting on the body equals zero. In such problems, both the translational forces and the rotational effects must balance, ensuring that there is no linear or angular acceleration.

Transcript

00:01 For this problem on the topic of equilibrium and elasticity, we are shown a 15kg block which is being held in place via pulley system.

00:08 A person's upper arm, which is vertical and their forearm at an angle of 30 degrees with the horizontal, is doing the work in holding the block in place.

00:20 We know the mass of the forearm and the hand together, as well as the center of mass distance, 15 centimeters, from the contact point of the foreman.

00:30 Bone and the upper arm bone or humerus.

00:33 We know the triceps will pull vertically upward of the forearm 2 .5 centimeters behind that contact point and the distance d3 is 35 centimeters.

00:46 We want to calculate the magnitude and direction of the force that the triceps exert on the forearm as well as the magnitude and direction that the humorous applies on the forearm.

01:00 Now all the forces here are vertical and all distances are measured along an axis inclined at 30 degrees.

01:08 Thus any trigonometric factor will cancel out and the application of talks about the contact point gives us that the force exerted by the triceps which we'll call ft is simply 15 kg times g which is 9 .8 meters per square second minus 2kg times 9 .8 meters per square second times 15 centimeters.

01:47 The first term needs to be multiplied by a distance of 35 centimeters to get the torque and all of this divided by 2 .5 centimeters.

02:04 And so calculating we get the force exerted by the tricep to be 1 .9 times 10 to the power 3, newton's...

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