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or does it depends on something else also?

2006-10-11 14:35:26 · 5 answers · asked by hypnonebula 3 in Science & Mathematics Physics

5 answers

The mass. The friction force is calculated using N*coefficient of friction. N= mg cos @. So, if the mass increase, the friction will also increase. The area will not influence the frictional force.

2006-10-11 14:47:13 · answer #1 · answered by huzaifahar2000 2 · 0 0

F=µN

F = frictional force
µ = coefficient of friction
N = Normal reaction force on the surface.

Greater mass will increase the normal reactive force, which will in turn increase the friction force.
Higher coefficient of friction on the surface will also increase the friction force.

The slope and direction the object is being dragged can also impact the frictional force because it also affects the normal (N).

Surface area theoretically should not affect the friction force. The greater the surface area the greater the chance of having a non-uniform coefficient of friction. In many real life circumstances this is the case and it becomes hard to determine the exact frictional force, but for theoretical understanding surface area should be disregarded.

2006-10-11 15:35:49 · answer #2 · answered by Anonymous · 0 0

It depends on the application. For an aircraft for instance, surface area definitely has an impact on the total amount of friction. When dealing with fluids, the amount of surface exposed will have an impact. When dealing with dragging a block across the floor, the only things that matter are the mass and the coefficient of friction (static or kinetic as the case may be) and the angle of the normal force relative to the angle of movement, as mentioned earlier.

2006-10-11 15:22:12 · answer #3 · answered by zmonte 3 · 0 0

both, the area creates a greater "drag" patch for the object to contact the surface with, and a greater mass creates more weight to drag.

2006-10-11 14:37:29 · answer #4 · answered by kevvsworld 3 · 0 0

Either will do it

2006-10-11 14:37:03 · answer #5 · answered by arbiter007 6 · 0 0

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