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

[Fusilli_Jerry89]

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?

 

[EthanB]

Start by posting your current work on the problem.

 

[Fusilli_Jerry89]

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]

How'd you get those numbers? Can you be a little more clear?

 

[Fusilli_Jerry89]

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.

 

[EthanB]

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.

 

[Fusilli_Jerry89]

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.

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

 

[EthanB]

If friction is acting down, we should be subtracting 42.6.

 

[Fusilli_Jerry89]

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?

 

[Fusilli_Jerry89]

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.

 

[EthanB]

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!

 

[Fusilli_Jerry89]

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?

 

[EthanB]

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.

 

[Fusilli_Jerry89]

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

 

[EthanB]

If the x direction is parallel to the slope, Fx means the forces parallel to the slope.

I mean force applied by Fa.