Solving Cart Force Question: Magnitude & Direction

[reminiscent]

Homework Statement

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?

Homework Equations

F=ma

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?

 

[cnh1995]

I don't understand what the question means when it states that "the child goes away." I would like to have some clues.

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.

 

[reminiscent]

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.

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.

 

[cnh1995]

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.

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

 

[cnh1995]

, but am I supposed to use the direction of the force that the baby exerts in the original problem?

Yes.

 

[reminiscent]

Yes.

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?

 

[cnh1995]

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?

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?

 

[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?

 

[reminiscent]

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?

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

 

[cnh1995]

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

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

 

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

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

 

[reminiscent]

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

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?

 

[cnh1995]

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?

Exactly.

 

[reminiscent]

Exactly.

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?

 

[cnh1995]

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?

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.

 

[reminiscent]

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.

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)

 

[cnh1995]

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)

Right!

 

[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)

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

 

[cnh1995]

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

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.

 

[reminiscent]

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.

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

 

[cnh1995]

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

You're welcome..

 

[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.

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.