Am I missing anything in this box problem?

[tibug]

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.

 

[voko]

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?

 

[tibug]

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.

 

[voko]

Your reasoning is entirely correctly. So sleep well :)

 

[Nickie]

Hello! Welcome to the community.

In response to your question, it seems like you have solved the problem correctly. You have correctly accounted for the y-component of the third force and assumed that the x-component does not matter on a frictionless surface. Your reasoning is sound and your calculations are correct.

However, it is always good to double check your work and make sure all assumptions are valid. In this case, you may want to consider if there are any other external forces acting on the box that could affect its movement. Additionally, it may be helpful to draw a free body diagram to visually see all the forces acting on the box.

Overall, your solution to the problem is reasonable and accurate. Great job!