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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?

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In summary, the work required to move the arm from position 1 to position 2 is -0.21mg(cos(30)-1), where m is the mass of the arm and g is the acceleration due to gravity. To find the force applied to move the arm, we use the equation W = F*delta x*cos(angle). By treating the arm as a single point mass attached to a rigid, massless rod, we can use the line integral to find the work done by gravity. Using trigonometry, we can also determine the exact height the arm has moved, making the calculation more accurate. Thank you Quasar987 and AM for your helpful explanations.

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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?

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Haroon Pasha said:

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

[tex]W = \int_{C}\vec{F}\cdot d\vec{r}[/tex]

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

[tex]x(\theta) = 0.21 sin(\theta)[/tex]

[tex]y(\theta) = -0.21 cos(\theta)[/tex]

[tex]0\leq \theta \leq 30°[/tex]

[tex]\vec{r}(\theta) = 0.21 sin(\theta)\hat{x} -0.21 cos(\theta)\hat{y}[/tex]

[tex] \vec{r'}(\theta) = 0.21 cos(\theta)\hat{x} + 0.21 sin(\theta)\hat{y}[/tex]

[tex]\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) [/tex]

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

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

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?

AM

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0.21 - h = 0.21 cos(30°)

Work is defined as the force applied to an object multiplied by the distance the object moves in the direction of the force. It is represented by the equation W = F * d.

To calculate work, you must first determine the force applied to the object and the distance it moves in the direction of the force. Then, you can use the equation W = F * d to find the work done.

The unit of measurement for work is the joule (J). This unit is derived from the basic units of mass (kilograms), length (meters), and time (seconds).

The angle of the force affects the work done because work is only done when the force is applied in the same direction as the movement of the object. If the force is applied at an angle, only the component of the force in the direction of the movement will contribute to the work done.

The work done to move an object from one position to another can be affected by various factors such as the magnitude and direction of the force applied, the distance the object moves, and the angle of the force. Other factors such as friction, air resistance, and the mass of the object can also affect the work done.

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