# Questions on force equations

1. Mar 4, 2006

### mrserv0n

Hello everyone, I have a physics test coming up and am working on the Pre-test however I have run into a problem and cannot find the proper equations anywhere to solve this...

Box B has a mass of Mb=35.2kg and moves with a constant velocity along the x axis. The track has a friction coefficent of U(friction sign)k =0.320.

draw a free body diagram and the four forces acting on it, Pulling force (Fp), Friction Force(Ff), Normal force (N) and finally Weight force (W). Also calculate the values of such forces.

Ok I found out that Friction force = U(friction sign)*Normal force

but I cannot find the equation to find weight force, normal force, or pulling force, What are they please? Im not asking for the solution to the problem just the equations used to solve it. Thanks for anyones time.

2. Mar 4, 2006

### BananaMan

well, lets look at this, you have a mass and you know the acceleration due to gravity, what equation links an acceleration a mass and a force? (this is actually the only formula you need)

you then should be able to work out the normal force knowing the box isn't passing through the surface.

from this you can work out the frictional force, and then we know because the box is moving at a constant velocity that all the forces are in balance, draw your diagram and take a look at what you think the pulling force should be

didn't want to do it all for you but hope it has helped

3. Mar 5, 2006

### andrevdh

The weight force is he force with which the earth attracts the box. According to Newton's universal gravitational law this amounts to
$$W=G\frac{Mm}{r_E^2}$$
where $G=6.67\times10^{-11}\ N.m^2/kg^2$ the universal gravitational constant. $M=5.98\times10^{24}\ kg$ is the mass of the earth and $r_E=6.37\times10^6\ m$ the radius of the earth. $m$ is the mass of the box in this case. In the above equation the terms
$$G\frac{M}{r_E^2}$$
comes to the same amount for all objects on the surface of the earth (well more or less due to some minor complicating factors). If you check the units of this calculation
$$\frac{N.m^2.kg}{m^2.kg^2}=\frac{N}{kg}=\frac{kg.m}{s^2.kg}$$
which comes to the units of an acceleration. This acceleration quantity is normally indicated with the symbol $g$ in physics and called the acceleration due to gravity. The weight force therefore becomes
$$W=gm$$
for any object on the surface of the earth. What do you get the value of $g$, the gravitational acceleration?

Last edited: Mar 5, 2006
4. Mar 5, 2006

### BananaMan

i was working under the assumption you was allowed to use the accepted value of acceleration due to gravity, if you arn't you would have to use this, but if not the problem is fairly simple, just look up the value in your book of constants (if you can't remember it)

5. Mar 5, 2006

### mrserv0n

thanks so much for a helpful answer it is exactly what I was looking for. Very much appreciated thank you again