# Kinetic frction between the hamburger and the ramp

1. Nov 4, 2004

### bballgirlweez

physics help plzzzzzz

Suppse a giant hamburger slides down a rampt aht has a 45 degree incline. The coefficient of kinetic frction between the hamburger and the ramp is .597, so that the net force actiong on the hamburger is 6.99 x 10^3 N. What is the mass of the hamburger? What is the magnitude of teh normal force that the ramp exerts on the hamburger?

2. Nov 5, 2004

### cepheid

Staff Emeritus
Work 'Backwards'

Instead of finding the vector sum of the various forces acting on the body, you are given the resultant force and asked to resolve it into all of its various components. So, draw the FBD and work backwards from what you know. You know the applied force and the angle, so you can resolve it into components perpendicular and parallel to the incline. The perpendicular component is the normal force. The parallel component is the net force on the burger in the direction down the ramp. What contributes to it? The // component of the weight. What takes away from it? The friction force. Find the latter, use it find the former. You can then solve for the weight.

3. Nov 6, 2004

### jai6638

just for my info, is 1194 kg the mass of it??

4. Nov 6, 2004

### jai6638

coz i did Ff=u+fn
6990=.597+ 9.8m ( since fn = mass x gravity )
6990/.597=9.8 m
m = 1194....

is that correct?

5. Nov 6, 2004

### cepheid

Staff Emeritus
Sorry, I don't understand what you are doing there. Could you maybe explain where the formula came from? Thanks!

6. Nov 6, 2004

### Skomatth

bball girl you need to show some of your work first.

Jai remember that:

F_f=uN

6.99 x 10^3 is the resultant force, not the force of friction.

7. Nov 7, 2004

### jai6638

well thats a formula i learnt..

Ff=Uk + Fn ... ( Force of friction = Coefiicient of Kinetic friction x Normal Force )

F_f= uN

can u please tell me what the variables in the above equation stand for?

is f_f - frictional force?
u - coefficient
N = normal force?

Thanks much for ur assistance..

after reading the question again, this is my thought process:

Fnet= 6990 N
Uk = .597

the following are the equatoins i have actually learned

F = ma
Fapp-Ff=ma
Ff=uk+Fn

i cant seem to substitute these values in the above equations to get the answer .. maybe if i only can find the acceleration so that i could substitute it in the F=ma equation to get the mass....

DAMN! i've done this chapter last year and am again doing it this year and i still cant get a simple problem.. :'(

8. Nov 7, 2004

### Windwaker2004

There is a way to get the acceleration of the hamburger...

Since $$\ F_a_p_p - F_f = ma$$

and

$$\ F_a_p_p = sin45(9.8m)\ and\ F_f = 0.597(9.8m)$$

The applied force was found from the parallel component of the force of gravity acting on the hamburger.

So...

$$\ F_a_p_p\ - F_f = ma$$

Sub in the values...

$$sin45(9.8m)\ -\ 0.597(9.8m) = ma$$

Now notice that the masses cancel out and you are left with acceleration. Solve for acceleration and sub that into the $$\ F_n_e_t = ma$$ equation.

The answer I came up with was 6472.2 kg

I haven't done this in a year or so, it would be good if someone could check it.

9. Nov 7, 2004

### jai6638

hmm i think your solution might be correct but its on an incline so there has to be a horizontal component of the force app too i think...

10. Nov 8, 2004

### Windwaker2004

Well, the $$\ sin45m$$ is the component of the force of gravity acting parallel to the ramp. The only other component is perpendicular to the ramp which creates the friction. The component of the force parallel to the ramp is the only one that needs to be looked at for force applied (remember that the force applied comes from gravity).

11. Nov 8, 2004

### Phymath

Suppse a giant hamburger slides down a rampt aht has a 45 degree incline. The coefficient of kinetic frction between the hamburger and the ramp is .597, so that the net force actiong on the hamburger is 6.99 x 10^3 N.
What is the mass of the hamburger?
$$ma = F_g + N + F_f$$
$$6.99 * 10^3 N = |{(mg sin 45 + -umg)\vect{i} + (mg - mg cos 45)\vect{j}}|$$
$$6.99*10^3 N = \sqrt{ (mg(sin 45 - u))^2 + (mg(1-cos 45))^2} = .312906mg$$
$$6.99*10^3 N/ (.312906*9.81 \fract{m}{s^2}) = m = 2276.77 kg$$

What is the magnitude of teh normal force that the ramp exerts on the hamburger?
N = mg (in the axis lines parrellel with the plane F.O.R.)
N = mg cos 45 (in the base of the ramp as the axis F.O.R.)

Last edited: Nov 8, 2004