# Cross product question find the magnitude of the force

loganblacke
cross product question.. "find the magnitude of the force"

## Homework Statement

A wrench 0.9 meters long lies along the positive y-axis, and grips a bolt at the origin. A force is applied in the direction of <0, 3, -2> at the end of the wrench. Find the magnitude of the force in newtons needed to supply 100 newton-meters of torque to the bolt.

## Homework Equations

|T| = |r x F| = |r||F|sin(theta)

## The Attempt at a Solution

I started with the point at the end of the wrench.. (0,.9,0) and then used the point given in the problem. (0, 3, -2) - (0, .9, 0) = (0, 2.1, -2). Is this the vector F?? I though it was, divided it by its magnitude to get the unit vector, but this doesn't seem to help as I think I need to find theta and then solve for |F|. Please help..

loganblacke

## Homework Statement

A wrench 0.9 meters long lies along the positive y-axis, and grips a bolt at the origin. A force is applied in the direction of <0, 3, -2> at the end of the wrench. Find the magnitude of the force in newtons needed to supply 100 newton-meters of torque to the bolt.

## Homework Equations

|T| = |r x F| = |r||F|sin(theta)

## The Attempt at a Solution

I started with the point at the end of the wrench.. (0,.9,0) and then used the point given in the problem. (0, 3, -2) - (0, .9, 0) = (0, 2.1, -2). Is this the vector F?? I though it was, divided it by its magnitude to get the unit vector, but this doesn't seem to help as I think I need to find theta and then solve for |F|. Please help..

I tried setting vector r as (0, .9, 0) and vector F as (0, 3, -2). I took the magnitude of vector r cross vector F which was the sqrt(-1.8^2) so 1.8. I found the magnitude of r to be .9 and the magnitude of F to be 3.6. I plugged it into the equation...

sin(theta) = |r x F|/(|r||F|)

... and found theta = 33.74. I plugged this into the first equation I mentioned in the problem statement and solved for the magnitude F but the answer was still wrong, my work looked like this....

100 = (.25)|F|*sin(33.74) soooo |F| = 400/sin(33.74) = 720.169

Mentor

## Homework Statement

A wrench 0.9 meters long lies along the positive y-axis, and grips a bolt at the origin. A force is applied in the direction of <0, 3, -2> at the end of the wrench. Find the magnitude of the force in newtons needed to supply 100 newton-meters of torque to the bolt.

## Homework Equations

|T| = |r x F| = |r||F|sin(theta)

## The Attempt at a Solution

I started with the point at the end of the wrench.. (0,.9,0) and then used the point given in the problem. (0, 3, -2) - (0, .9, 0) = (0, 2.1, -2). Is this the vector F??
No. All you know about F is the direction in which it is applied. You know |T|, |r|, and you should be able to find theta, and from these you can solve algebraically for |F|.
I though it was, divided it by its magnitude to get the unit vector, but this doesn't seem to help as I think I need to find theta and then solve for |F|. Please help..

loganblacke

No. All you know about F is the direction in which it is applied. You know |T|, |r|, and you should be able to find theta, and from these you can solve algebraically for |F|.

Marc, I figured that part out, I added my work to my original post. I am doing something wrong when finding theta, can you take a look please.

Staff Emeritus
Homework Helper
Gold Member

I tried setting vector r as (0, .9, 0) and vector F as (0, 3, -2). I took the magnitude of vector r cross vector F which was the sqrt(-1.8^2) so 1.8. I found the magnitude of r to be .9 and the magnitude of F to be 3.6. I plugged it into the equation...

sin(theta) = |r x F|/(|r||F|)

... and found theta = 33.74. I plugged this into the first equation I mentioned in the problem statement and solved for the magnitude F but the answer was still wrong, my work looked like this....

100 = (.25)|F|*sin(33.74) soooo |F| = 400/sin(33.74) = 720.169
The usual way to do this is to construct a unit vector in the direction of (0,3,-2), then the force vector is the magnitude of F times the unit vector.

Let's try a way closer to what you attempted:

$$\text{Let }\vec{F}=k(0,\,3,\,-2)=(0,\,3k,\,-2k)\,,$$ where k is a scalar constant.

Then the torque is given by $$\vec{\tau}=\vec{r}\times\vec{F}=(0,\,0.9,\,0)\times(0,\,3k,\,-2k)\,.$$

Can you take it from here?

loganblacke

The usual way to do this is to construct a unit vector in the direction of (0,3,-2), then the force vector is the magnitude of F times the unit vector.

Let's try a way closer to what you attempted:

$$\text{Let }\vec{F}=k(0,\,3,\,-2)=(0,\,3k,\,-2k)\,,$$ where k is a scalar constant.

Then the torque is given by $$\vec{\tau}=\vec{r}\times\vec{F}=(0,\,0.9,\,0)\times(0,\,3k,\,-2k)\,.$$

Can you take it from here?

We actually haven't done anything with a scalar constant that I'm aware of.. but when you say the usual way, do you construct the unit vector from the origin?

Staff Emeritus
Homework Helper
Gold Member

We actually haven't done anything with a scalar constant that I'm aware of.. but when you say the usual way, do you construct the unit vector from the origin?
No. The vector (0,3,-2) has a magnitude of √(13). So, a vector of magnitude = 1, in the direction of (0,3,-2), can be written as (1/√(13)(0,3,-2), (so you still need to multiply a vector by a scalar). At any rate, then the force vector F is given by F=|F|(1/√(13)(0,3,-2).

Another way to look at this is that the magnitude of the torque is equal the component of F that's perpendicular to r, times the magnitude of r. Since (0,3,-2) only has y & z components, it's the z component of F that's ⊥ to r.

|r|·Fz=100 newton·meters. Solve for Fz.

The ratio of Fz/Fy = -2/3.

Last edited:
loganblacke

No. The vector (0,3,-2) has a magnitude of √(13). So, a vector of magnitude = 1, in the direction of (0,3,-2), can be written as (1/√(13)(0,3,-2), (so you still need to multiply a vector by a scalar). At any rate, then the force vector F is given by F=|F|(1/√(13)(0,3,-2).

I understand that the vector <0, 3, -2> has a magnitude of sqrt(13). That was one of the first things that I did. And the unit vector is the vector/magnitude.. get that. But if you don't have a value for vector F, I don't see how you can find the magnitude of F to answer the question. In your equation you have the vector F = |F|*unit vector, but is it the unit vector of F? If so then the magnitude of vector F would be sqrt(13), which would be the answer.. Totally not following you.

Zebrostrich

I understand that the vector <0, 3, -2> has a magnitude of sqrt(13). That was one of the first things that I did. And the unit vector is the vector/magnitude.. get that. But if you don't have a value for vector F, I don't see how you can find the magnitude of F to answer the question. In your equation you have the vector F = |F|*unit vector, but is it the unit vector of F? If so then the magnitude of vector F would be sqrt(13), which would be the answer.. Totally not following you.

Perhaps you're misunderstanding a concept behind a unit vector. First of all, the unit vector always has a magnitude of one. It purely specifies direction.

Now think: you are given a vector that specifies the direction of the force, but does not have a magnitude of 1. What you want is a vector F in the same direction as <0, 3, -2>, but with a magnitude of |F|. That is the purpose of calculating the unit vector. Once you obtain the unit vector, you compute the vector F (symbolically, of course, since the magnitude is unknown) by multiplying the unit vector by |F|. From there you use what you know about the cross product and torque to set up an equation to solve for |F|.

(Hint: you should not need theta)