Cross product question find the magnitude of the force

  • #1
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..
 

Answers and Replies

  • #2


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
 
  • #3
34,867
6,600


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..
 
  • #4


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.
 
  • #5
SammyS
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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:

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

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

Can you take it from here?
 
  • #6


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:

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

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

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?
 
  • #7
SammyS
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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:
  • #8


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.
 
  • #9


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)
 

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