Solving Cart Force Question: Magnitude & Direction

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In summary, the child should exert a force of 17 N in the direction of 90 degrees. The weight of the cart is 840 N and the box moves at 2.0 m>s2 in the +x-direction.
  • #1
reminiscent
131
2

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?
vV6kL4g.png


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?
 
Last edited:
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  • #2
reminiscent said:
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.
 
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  • #3
cnh1995 said:
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.
 
  • #4
reminiscent said:
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.
 
  • #5
reminiscent said:
, but am I supposed to use the direction of the force that the baby exerts in the original problem?
Yes.
 
  • #6
cnh1995 said:
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?
 
  • #7
reminiscent said:
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?
 
  • #8
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
cnh1995 said:
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?
 
  • #10
reminiscent said:
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.
 
  • #11
cnh1995 said:
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.
 
  • #12
cnh1995 said:
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?
 
  • #13
reminiscent said:
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.
 
  • #14
cnh1995 said:
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?
 
  • #15
reminiscent said:
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.
 
  • #16
cnh1995 said:
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)
 
  • #17
reminiscent said:
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!
 
  • #18
reminiscent said:
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?
 
  • #19
reminiscent said:
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.
 
  • #20
cnh1995 said:
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
 
  • #21
reminiscent said:
Thank you so much for your help! I appreciate it. :D
You're welcome:smile:..
 
  • #22
cnh1995 said:
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.
 

Related to Solving Cart Force Question: Magnitude & Direction

What is the formula for calculating cart force?

The formula for calculating cart force is Fc = m x a, where Fc is cart force, m is mass, and a is acceleration.

How do you determine the magnitude and direction of cart force?

To determine the magnitude of cart force, you can use the formula Fc = m x a. The direction of cart force can be determined by the direction of the acceleration or by using the right-hand rule.

What is the difference between cart force and net force?

Cart force refers to the force applied specifically to the cart, while net force is the sum of all forces acting on an object. In the context of solving cart force questions, net force is important because it helps determine the acceleration of the cart.

How does mass affect cart force?

The mass of an object does not directly affect the cart force, but it does affect the acceleration of the cart. The greater the mass, the more force is needed to accelerate the cart at a certain rate.

What are some common mistakes to avoid when solving cart force questions?

Some common mistakes to avoid when solving cart force questions include not considering all the forces acting on the cart, using the wrong formula or direction for cart force, and not paying attention to units. It is also important to make sure the acceleration and force are in the same direction when using the formula Fc = m x a.

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