# Urgent Friction help!

1. Does a graph of force of friction vs force normal pass through the origin? Explain.

2. Does the force of kinetic friction depend on the weight of the object? Explain.

3. Does the coefficient of kinetic friction depend on the weight of the object? explain

This is not on an icline plane.

Fuego
1. yes.

F = &mu;R

&mu; does not = 0

where F = friction, R = normal reaction force and &mu; = coefficient of friction.

if F = 0, then R must also equal 0 to satisfy the above equation. therefore a graph of F against R would pass through the origin.

2. yes.

for an object moving on a horizontal plane, R = mg (where mg = weight). therefore, for an object moving on a horizontal plane:

F = &mu;mg

3. no.

F is proportional to R and hence F is proportional to mg. &mu; is just the constant of proportionality. it is a property of the surface(s) experiencing friction. it is dependant upon the shape and material of the surfaces.

ummm i don;t get what those things mean...

what is R?

Can you explain it in a simpler way?

The equations given to me are

ì = Ff/Fn

Fnet = mg

Fuego
in the equations given to you, what are i, Ff and Fn? you need to state what the symbols are representing.

in the equation i used, F is the frictional force acting on an object moving on a surface, R is the normal reaction force on the object and &mu; is the coefficient of friction between the object and the surface.

the frictional force F on the object is proportional to the normal reaction force R acting on the object:

F is proportional to R.

to turn this into an equation, a constant of proportionality is required. this is &mu;, the coefficient of friction:

F = &mu;R

&mu; cannot = 0. if &mu; = 0 there would be no friction.

in a graph of F = &mu;R, F would be the vertical axis and R would be the horizontal axis. &mu; would be the gradient of the graph. if the graph passes through the origin, then F must = 0 when R = 0. put R = 0 into the equation:

F = &mu;0
F = 0

so F does = 0 when R = 0 therefore the graph passes through the origin.

oop i meant u = Ff/Fn u = coefficent of friction, Fn = Force Normal, Ff = Force of friction.

How does the surface area of the block affect the force of friction or the coefficient of friction?

renedox
Well the surface area will affect the force of friction because quite obviously, the more there is, the more it has to fight to get from A to B

Fuego
kangta, the equation you are using is exactly the same as mine only the symbols are different.

you have used Ff for frictional force, and i have used F. you used Fn for normal reaction force, i used R. you used u for coefficient of friction, i used &mu;.

u = Ff/Fn can be rearanged to Ff = uFn, which is in the same form as i used.

the area in of the block in contact with the surface it is moving on affects the coefficient of friction u. u is a function of the area of the block in contact with the surface, and the material of the block.

Mentor
Originally posted by kangta

How does the surface area of the block affect the force of friction or the coefficient of friction?
Coeficient of friction includes surface area and surface area relates pressure and force. So if you increase surface area, the pressure decreases (for a given force) and the 'amount of surface to grab' increases by the same amount. So the friction force stays the same.

I'm not sure if that made sense, so let me try a mathematical explanation: if you break out a surface area term from the coefficient of friction, you also have to divide force by surface area (getting pressure). The two surface area terms cancel out.

Fuego
this has nothing to do with pressure.

the coefficient of friction &mu; is a function of the contact area between the two surfaces amongst other things.

F = &mu;R

&mu; = f(A) where f(A) is a function of the contact area. thus by changing A you change &mu;. by changing &mu; you change the frictional force for a given normal reaction force.

you can test this by experiment. two objects of identical mass but with different contact areas moving on a horizontal plane will experience different frictional forces.

this is why, for example, wide tyres give more grip than narrow tyres.

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