What is the maximum force to push a 50 kg wagon with a 15 kg box on an incline?

  • Thread starter Fusilli_Jerry89
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In summary: So if we are pushing the wagon up the slope, we need to supply an extra 4.6 newtons to the downward force of friction.
  • #1
Fusilli_Jerry89
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a 50 kg wagon is being pushed up a 15 degree incline. The coefficient of friction between the wagon and the slope is 0.15. A 15 kg box sits on top of the wagon and the coefficient between the wagon and the box is 0.30. What is the maximum force that you can push the wagon up the slope with before the box starts to slide?
 
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  • #2
Start by posting your current work on the problem.
 
  • #3
well first i drew it and came up with:
F-127-71+42.6=50a for the wagon and concluded that an extra 4.6 N must be directed down the slope in order to move the box. I am clueless now.
 
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  • #4
How'd you get those numbers? Can you be a little more clear?
 
  • #5
lol too much typing ok here I go
the 127 is the force of gravity acting on the big box down the slope, the 71 is the force of friction between the big box and the slope the 42.6 is the force of friction acting up on the wagon.
 
  • #6
Sum the forces on the block in one free-body diagram, then sum the forces on the cart in another.

There should be no force of friction acting up the ramp on the wagon. The wagon's motion is upwards, and friction resists the direction of motion.

Once we have four equations from summing our forces, find a way to relate the maximum static friction between the block and cart to the force being applied to the cart. Plug in the maximum friction and get the maximum force.
 
  • #7
Fusilli_Jerry89 said:
well first i drew it and came up with:
F-127-71+42.6=50a for the wagon and concluded that an extra 4.6 N must be directed down the slope in order to move the box. I am clueless now.

EthanB said:
Sum the forces on the block in one free-body diagram, then sum the forces on the cart in another.

There should be no force of friction acting up the ramp on the wagon. The wagon's motion is upwards, and friction resists the direction of motion.

Once we have four equations from summing our forces, find a way to relate the maximum static friction between the block and cart to the force being applied to the cart. Plug in the maximum friction and get the maximum force.

I put the force of friction acting down
 
  • #8
If friction is acting down, we should be subtracting 42.6.
 
  • #9
i am only able to come up with 2 equations: F-155.4=50a for the wagon and F=4.6 for the box. What else is there?
 
  • #10
isnt the 42.6 N acting upwards tho? bcuz the box isn't moving and the friction has to be acting upwards to keep it from moving.
 
  • #11
The friction between the box and wagon acts in different directions on each object. It acts downwards on the wagon and upwards on the box.

We can sum the forces up/down the ramp for each object, then perpendicular to the ramp for each object. Maybe you already took that into account and found numerical answers for everything. I haven't plugged the numbers and compared them to what you've gotten. My carpel tunnel is acting up and I'm going to sign off for the night. Good luck!
 
  • #12
What I don't get is how can you supply extra Newtons downwards without actually touch the box? The force of friction stays the same, this has to do with inertia but I have no idea how to solve for this? Maybe I missed something in Physics last year, could anyone help?
 
  • #13
Fusilli_Jerry89 said:
What I don't get is how can you supply extra Newtons downwards without actually touch the box? The force of friction stays the same, this has to do with inertia but I have no idea how to solve for this? Maybe I missed something in Physics last year, could anyone help?

I'm not sure what extra Newtons you're referring to. Lay out to me what you have and I will go from there. Try to be as specific as possible. Use a format such as:

Fx(on wagon) = Fa - Ff(ramp) ... etc.
 
  • #14
Sorry I don't understand that format, is Fx sposed to be parallel to the slope? and what is Fa?

For the wagon:
Ff = 70.95 N
Fg = 127 N
F = ?

For the box:
Ff = 42.6 N
Fg = 38 N
 
  • #15
If the x direction is parallel to the slope, Fx means the forces parallel to the slope.

I mean force applied by Fa.
 
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Related to What is the maximum force to push a 50 kg wagon with a 15 kg box on an incline?

1. What is the Box Sliding Wagon problem?

The Box Sliding Wagon problem is a classic physics problem that involves a box resting on a frictionless, horizontal surface and attached to a wagon. The wagon is initially at rest, and when a force is applied to the wagon, it starts moving with the box on top of it. The question is, what happens to the box when the wagon stops?

2. What are the key principles involved in the Box Sliding Wagon problem?

The Box Sliding Wagon problem involves principles of Newton's Laws of Motion, specifically the first and second laws. The first law states that an object at rest will remain at rest unless acted upon by an external force, and an object in motion will remain in motion with a constant velocity unless acted upon by an external force. The second law states that the net force on an object is equal to the mass of the object multiplied by its acceleration.

3. How is friction involved in the Box Sliding Wagon problem?

Friction plays a crucial role in the Box Sliding Wagon problem. In the real world, there is no such thing as a perfectly frictionless surface, so in this problem, we assume that the surface has no friction. However, the box and wagon are still in contact with each other, creating a frictional force. This force acts in the opposite direction of the motion and can affect the acceleration and final position of the box on the wagon.

4. What is the equation used to solve the Box Sliding Wagon problem?

The equation used to solve the Box Sliding Wagon problem is F = ma, where F is the net force, m is the mass of the box, and a is the acceleration. This equation allows us to calculate the force and acceleration acting on the box and wagon system and determine the resulting motion.

5. What are some real-world applications of the Box Sliding Wagon problem?

The Box Sliding Wagon problem has real-world applications in the design of transportation systems, such as trains and roller coasters. It also has applications in understanding the motion of objects on inclined planes and the effects of friction. Additionally, the principles involved in this problem are essential in the development of technologies such as self-driving cars and robotics.

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