What force is needed to keep the box at rest on an inclined plane?

In summary: If you have a horizontal force F, what's its component parallel to the incline (given the incline angle)?In summary, the problem involves a box placed on a frictionless inclined plane with an angle of 18° from the horizontal. The box has a mass of 35 kg and the question is asking for the magnitude of acceleration along the inclined plane and the magnitude of force needed to keep it at rest. The equation F=ma is used to solve for the acceleration, while the equation F=mg*sin is used to find the force needed to keep the box at rest. However, the force applied must be parallel to the incline, so the component of the force parallel to the incline must be equal to mg*sin
  • #1
map7s
146
0

Homework Statement



(a) A box is placed on an frictionless inclined plane with an angle of 18° from the horizontal. The box has a mass of 35 kg, what is the magnitude of the acceleration of the box along the inclined plane?
(b) What magnitude of force would you need to apply to the box in order to keep it at rest? Assume there is no friction and the force applied is parallel to the ground.


Homework Equations



F=ma
ma=mgsin

The Attempt at a Solution



I got the answer to the first part. I am just confused about the second part. I get that Fnet=0 N in order for the entire system to not move. However, I was thinking that, in order for that to happen, whatever the force would be to begin with, that would have to be the opposing force, as well (i.e. if F=mgsin then for Fnet=0, F=mgsin-mgsin=0, therefore F=mgsin in order for Fnet=0)...but that feels very confusing to me...I know that I have to go against gravity...but...I don't really know how to set up the equation for this problem...
 
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  • #2
Sounds to me like you are on the right track but just don't realize it. Gravity exerts a force down the incline equal to [itex]mg\sin\theta[/itex]. To cancel that force, you need to exert the same force up the incline. Done! (Just compute the actual value of the force.)
 
  • #3
okay...that was what I was thinking...but would it actually be the value of mgsin? (because that was what I originally thought and it turns out that I was wrong...) would I have to use the acceleration that I found in the first part? maybe do something like mgsin-masin?...or would that just not work...
 
  • #4
Ah... I misread the problem. The force is applied parallel to the ground, not parallel to the incline. (Sorry about that.) That's why [itex]mg\sin\theta[/itex] is wrong.

You need a horizontal force whose component parallel to the incline equals [itex]mg\sin\theta[/itex].

If you have a horizontal force F, what's its component parallel to the incline?
 
  • #5
...I think I understand what you are saying, but I'm not sure about setting up the equations...
it can't be just mgcos because, even though that is a horizontal force it doesn't equal to mgsin...right?
 
  • #6
Start by answering this question: If you have a horizontal force F, what's its component parallel to the incline (given the incline angle)?
 
  • #7
...would it be mgsin?
 
  • #8
Forget about the details of this problem, just answer the question in general for a force F.
 
  • #9
would it be Fsin?
 
  • #10
map7s said:
would it be Fsin?
No. (If F were a vertical force like gravity, then that would be true.)
 

1. What is net force?

Net force is the overall force acting on an object, taking into consideration all of the individual forces acting on it. It is calculated by adding or subtracting all of the forces acting on an object in a specific direction.

2. How is net force related to keeping a box at rest?

In order for an object, such as a box, to remain at rest, the net force acting on it must be zero. This means that all of the forces acting on the box must be balanced and cancel each other out.

3. What types of forces can act on a box to keep it at rest?

There are several types of forces that can act on a box to keep it at rest. These include gravitational force, normal force, and frictional force. Each of these forces must be balanced in order for the box to remain at rest.

4. Why is it important to understand net force when keeping a box at rest?

Understanding net force is important because it allows us to predict and control the motion of objects. In the case of keeping a box at rest, knowing the net force acting on it helps us determine what forces are needed to balance it and keep it from moving.

5. How can the net force be calculated to keep a box at rest?

The net force can be calculated by adding or subtracting all of the individual forces acting on the box in a specific direction. If the net force is zero, then the box will remain at rest. If the net force is not zero, then the box will accelerate in the direction of the net force.

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