# Homework Help: Cart Force question

1. Feb 7, 2016

### reminiscent

1. The problem statement, all variables and given/known data
Two adults and a child want to push a wheeled cart in the direction marked x in Fig. P4.33. The two adults push with horizontal forces F1 and F2 as shown, except that the child goes away. Keeping the magnitude of F1 and F2 the same, what angle should F1 make to have the box move in the desired direction? Using that angle, and again using the box’s weight of 840 N, what will the new acceleration be?

This problem is an edit of an actual problem, which have subquestions as follows:
(a) Find the magnitude and direction of the smallest force that the child should exert. Ignore the effects of friction. (answer = 17 N 90 degrees clockwise from x-direction)
(b) If the child exerts the minimum force found in part (a), the cart accelerates at 2.0 m>s2 in
the +x-direction. What is the weight of the cart?

2. Relevant equations
F=ma

3. The attempt at a solution
So I get that the new problem is focusing on the box. We are giving the weight of the box, but am I supposed to use the direction of the force that the baby exerts in the original problem?

Last edited: Feb 7, 2016
2. Feb 7, 2016

### cnh1995

In the original question, they are asking what should be the angle of 100N force so that the box keeps moving along the X axis.

Last edited: Feb 7, 2016
3. Feb 7, 2016

### reminiscent

Sorry, the original diagram had a 60 degree angle right there. Also, I am directing this question towards the edited problem, not the original problem.

4. Feb 7, 2016

### cnh1995

Ok. Here's a hint: The box moves only along the X direction, after the kid applies his force.

5. Feb 7, 2016

Yes.

6. Feb 8, 2016

### reminiscent

Okay, is this thought correct:
The weight of the box points down, therefore it is 840 N and it can be used as a component force, correct?

7. Feb 8, 2016

### cnh1995

I don't think so. Weight of the box is useful to find its mass and acceleration. Did you understand the hint I gave above?

8. Feb 8, 2016

### cnh1995

What can you say about the components of the forces acting on the box? It is moving in the x direction. What does this tell you?

9. Feb 8, 2016

### reminiscent

That it is not moving in the y-direction, therefore F=ma=0 in the y-direction, correct?

10. Feb 8, 2016

### cnh1995

Yes. The components in the y direction add up to 0.

11. Feb 8, 2016

### reminiscent

How does this help me for the x-direction, though? I think the picture is what is confusing me.

12. Feb 8, 2016

### reminiscent

Is it because you can add the y components of each force given in the diagram (like F2y, F1y), equal them to 0, and since you can find F2y, you can ultimately find F1y?

13. Feb 8, 2016

### cnh1995

Exactly.

14. Feb 8, 2016

### reminiscent

Does gravity need to be taken into account or no?
If not, I found F1y to be 70 N. To find the angle, it is sin-1(70/100) = 44.4 degrees.
How do I find the acceleration using the weight and the angle?

15. Feb 8, 2016

### cnh1995

Gravity is in the z direction.It is only useful for finding the mass of the box. By equating y components to 0, you will get the minimum force applied by the child.

16. Feb 8, 2016

### reminiscent

Oh, in this problem, I do not need to worry about the child. The initial problem was centered on the child, but the new problem asks how to find the acceleration of the box if the box was just being pushed.
So to find the mass of the box, do I do -840 N = m * (-9.8 m/s^2)? Then do I use that mass to find the acceleration of the box in the x-direction? (F2x+F1x = ma)

17. Feb 8, 2016

### cnh1995

Right!

18. Feb 8, 2016

### reminiscent

Also do I have to count the z-direction to be correct in this?

19. Feb 8, 2016

### cnh1995

No. Since the weight is in z direction, it will not affect x and y components. It is simply giving you the mass of the box.

20. Feb 8, 2016

### reminiscent

Thank you so much for your help! I appreciate it. :D

21. Feb 8, 2016

### cnh1995

You're welcome..

22. Feb 8, 2016

### cnh1995

The real reason why weight doesn't affect x and y components is because it is balanced by the normal reaction. If the box were free falling, all the three forces would have a combined effect on the box.