Calculating Work Done to Move Arm from Position 1 to 2

[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?

 

[quasar987]

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?

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).

 

[Andrew Mason]

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?

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

AM

 

[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.

 

[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°)