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Am I missing anything in this box problem?

  • Thread starter tibug
  • Start date
  • #1
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Hi guys! I'm new here.

Here is the problem, verbatim: Two people pull to the right on a box with forces of 100 newtons at an angle of 60 degrees with respect to the x axis into the first quadrant and 80 newtons at an angle of 20 degrees down into the fourth quadrant respectively. A third person is to also pull on the box to exert a force that when the three forces are combined, the box moves to the right along the x axis. What is the magnitude and direction of the MINIMUM force the third person should exert to make this happen?

My first source of confusion was this: I can find what the y-component of the third force is in order to zero any movement in the y-direction, but how do I know what x-component is required? I think I solved that problem by assuming that the problem assumes a frictionless surface, so the x-component of the force does not matter unless it is in the opposite direction. The forces on the box by persons 1. and 2. are in the x-direction anyways, so why would the third person need to exert a force in the x-direction? Is that reasonable or completely off?

With that assumption made, I made this equation:

V1y + V2y + V3y = 0

Substitute in the directions and magnitudes:

100N(sin60) + (-80N(sin30)) + (-V3y) = 0

Solve:

V3y = 46.602 N and (I assume V3x = 0)

When I go ahead and add x-components to find the vector of V1 + V2 + V3 = V4...

100(cos60) + 80(cos30) + 46.6(cos90) = 119.28

And then of course, V4 = √(119.28^2+0^2) = 119.28

So my V4 vector (the sum of the force on the box) is 119.28 N in the + x-direction.


I finished the problem but I feel like it's wrong, but I can't think of any reason why it would be.
 

Answers and Replies

  • #2
6,054
390
Note that you are required the MINIMUM force. Is your force such a minimum? Can you think of any force with a lesser or greater magnitude that the one you have found, which still satisfies the other requirements?
 
  • #3
4
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Thanks for the reply! I can't think of any force with a lesser magnitude that would satisfy the other requirements. The value of sine at 90 degrees is 1, so if I rotate V3 around the origin at all towards the x-axis, sine would decrease and the magnitude of the vector would have to be higher in order to provide 46.602N of force in the downward y-direction. And the magnitude in the x direction would increase. Of course, that means that I can think of forces of greater magnitude that would satisfy the other requirements. Any way I look at it, I can't think of a force less than 46 that would keep the box moving along the x axis.

I think I've got it right, but I may just need to sleep on it. That happened a lot with me in Calculus. I'd stay up late doing homework, get it done, then get up the next morning and I realize I did it wrong...I dream of Calculus...haha.

Again, thank you.
 
  • #4
6,054
390
Your reasoning is entirely correctly. So sleep well :)
 

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