# Angle between Fz and S, picture included

• gap0063
In summary, the angle between the body diagonal and the front-left edge of the block is θ=arcsin(99k*S/(mag(99k*S)))
gap0063

## Homework Statement

Hint: W = ~S · ~F .
Consider a force ~F with components Fx =28 N, Fy = 57 N, and Fz = 99 N, as illustrated in the figure below. Work is done when a particle moves up the vertical z-axis a distance of 4 m.
[PLAIN]http://img691.imageshack.us/img691/5594/problem4e.jpg
What is the angle θ between the body diagonal which represents the force ~F and the front-left edge of the block which represents both the z-component Fz of the vector ~F and the vector ~S the particle traversed?

## Homework Equations

They give me a hint of W= S *F
but I don't know S or W...

## The Attempt at a Solution

F= 28i + 57 j + 99 k

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Work for a constant force is the dot product between the force and the displacement,
$$W=\vec{F} \cdot \vec{d}$$.

What are the two definitions of the dot/scalar product, and how can you use them to find the angle θ?

jhae2.718 said:
Work for a constant force is the dot product between the force and the displacement,
$$W=\vec{F} \cdot \vec{d}$$.

What are the two definitions of the dot/scalar product, and how can you use them to find the angle θ?

W= F * d= mag(F delta d) costheta?
so is the distance only in the z direction?
and if so does that mean only Fzk matter?

The problem says "Work is done when a particle moves up the vertical z-axis a distance of 4 m", so what does that imply about the direction of the displacement?

You can relate the dot product in terms of the magnitudes and cos(theta) with the dot product in terms of the components to find cos(theta), and then use arcos(cos(theta)).

jhae2.718 said:
The problem says "Work is done when a particle moves up the vertical z-axis a distance of 4 m", so what does that imply about the direction of the displacement?

You can relate the dot product in terms of the magnitudes and cos(theta) with the dot product in terms of the components to find cos(theta), and then use arcos(cos(theta)).

So the direction is in the z direction... so does this mean I only need to care about Fzk(which is also S) and F?

Essentially; $$W=F_z\hat{k} \cdot S\hat{k}$$ in this case. Now, from this, how can you get $$\theta$$?

Note that you don't need to take $$\vec{F}$$ itself in the calculation once you've put it into component form.

jhae2.718 said:
Essentially; $$W=F_z\hat{k} \cdot S\hat{k}$$ in this case. Now, from this, how can you get $$\theta$$?

Note that you don't need to take $$\vec{F}$$ itself in the calculation once you've put it into component form.

So, Fzk=99k
W= 99k * S= mag (99k*S)costheta
theta=arccos( 99k*S)/mag (99k*S)

but how do I get S?

S is 4 m.

You're making a mistake. It is the magnitude of $$\vec{F}$$ in the dot product, so you should have:
$$99\,\mathrm{N}\cdot 4\,\mathrm{m}=|\vec{F}| \cdot 4\,\mathrm{m} \cos(\theta)$$.

$$|\vec{F}|=\sqrt{F_x^2+F_y^2+F_z^2}$$.

jhae2.718 said:
S is 4 m.

You're making a mistake. It is the magnitude of $$\vec{F}$$ in the dot product, so you should have:
$$99\,\mathrm{N}\cdot 4\,\mathrm{m}=|\vec{F}| \cdot 4\,\mathrm{m} \cos(\theta)$$.

$$|\vec{F}|=\sqrt{F_x^2+F_y^2+F_z^2}$$.

Oh, I see now.

Thanks so much!

## 1. What is the angle between Fz and S?

The angle between Fz and S can vary depending on the specific situation or context. In general, it refers to the angle between the direction of the force (Fz) and the direction of displacement (S). It is commonly measured in degrees or radians.

## 2. How do you calculate the angle between Fz and S?

The angle between Fz and S can be calculated using trigonometric functions such as sine, cosine, and tangent. It involves finding the ratio of the side opposite the angle (Fz) to the side adjacent to the angle (S) in a right triangle formed by these two vectors.

## 3. Why is the angle between Fz and S important?

The angle between Fz and S is important in understanding the relationship between force and displacement in a given situation. It can also help determine the direction of the resulting motion or the work done by the force.

## 4. Can the angle between Fz and S be negative?

Yes, the angle between Fz and S can be negative. This indicates that the force and displacement are in opposite directions, resulting in negative work being done.

## 5. How does the angle between Fz and S affect the magnitude of the force?

The angle between Fz and S does not affect the magnitude of the force. However, it does affect the component of the force in the direction of displacement, which is important in calculating the work done by the force.

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