# Work Problem

1. Apr 17, 2005

### Haroon Pasha

An arm has a mass of 7.0 kg. Treating the arm as if it were a single point mass m attached to a rigid massless rod, determine the work that must be done by the deltoid muscle to move the arm from position 1 to position 2. (Position one and two have an angle of 30 degrees between them and it is 21.0 cm from the shoulder to the elbow of the arm)

By doing a little trigonometry, I figured out that the distance between position 1 and position 2 is about 0.12m. I will use the equation
W(for work)=F*delta x*cos(angle)

How do I find the force applied to move it from position one to two?

2. Apr 17, 2005

### quasar987

Your arm is in the earth's gravitational field. You will have to do work against this force. The minimum work you have to do to move your arm 30° is the work done BY GRAVITY on your arm as it moves 30°. So let's find that. If we could find exactly what HEIGHT your arm had moved, it would be easy: W = -mgh. But we can't apparently. So we'll stick with the definition of work done by a force, which is the line integral

$$W = \int_{C}\vec{F}\cdot d\vec{r}$$

C is a portion of circle. Easy to parametrize. Let's do that..

$$x(\theta) = 0.21 sin(\theta)$$
$$y(\theta) = -0.21 cos(\theta)$$
$$0\leq \theta \leq 30°$$
$$\vec{r}(\theta) = 0.21 sin(\theta)\hat{x} -0.21 cos(\theta)\hat{y}$$
$$\vec{r'}(\theta) = 0.21 cos(\theta)\hat{x} + 0.21 sin(\theta)\hat{y}$$

$$\Rightarrow \int_{C}\vec{F}\cdot d\vec{r} = \int_0^{30°} \vec{F}(\vec{r}(\theta)) \cdot \vec{r'}(\theta)d\theta = \int_0^{30°}-0.21mgsin(\theta)d\theta = 0.21mg(cos(30)-1)$$

So your muscle will have to do work in the amount -0.21mg(cos(30)-1).

Last edited: Apr 17, 2005
3. Apr 17, 2005

### Andrew Mason

The force is mg. What is W in terms of mass, g, and the height?

AM

4. Apr 17, 2005

### Haroon Pasha

Wow. that makes sense. That was very helpful. Even though I am not in calc based physics, I still understoon your reasoning. Thanks Quasar987 and AM both.

5. Apr 17, 2005

### quasar987

It turns out we can get the height exactly directly from trigonometry. It would require a drawing but the key is

0.21 - h = 0.21 cos(30°)