11Yfriction


 * Friction **

Stationary Friction

An object sitting on a surface is stationary. If a force is attempting to push or pull the object sideways but the object is not moving, then there must be some other force resisting that motion.

The force resisting the motion is __Friction__.

Since the object is not moving, the friction force must be equal and opposite to the __component__ of the pushing/pulling force that is __parallel to the surface__. (by Newton's first law)

Example 1

A 10kg block is sitting on a horizontal surface. A rope attached to the block is being pulled up at 30º above the horizontal with a force of 4 Newton. Calculate the magnitude of the frictional force that is preventing the block from moving.

math \\ . \qquad Fr=4\cos30^\circ \\. \\ . \qquad Fr=2\sqrt{3} \; \; \text{Newtons} math
 * The block is not moving sideways so the horizontal forces must be equal and opposite. **

Sliding Friction

When an object is sliding along a surface, friction acts opposite to the direction of motion and parallel to the surface. The __magnitude__ of the frictional force is given by: math \\ Fr=\mu N \qquad \text{where} \\ . \qquad \qquad Fr= \text{ Friction force} \\ . \qquad \qquad N = \text{ Normal} \\ . \qquad \qquad \mu = \text{ (mu) the Coefficient of Friction -- constant -- no units} math

The value of m ( **mu)** will be different depending on what two surfaces are rubbing against each other. Mu** ( **m **)** is usually between 0 and 1.

For a given question, you will either be given a value for m, or you will be asked to calculate m from Fr = m N.

Example 2 A car with a mass of 900kg is being towed at a constant speed of 12 m/s. The tension in the tow-rope is 400 Newtons. Find the Friction Force acting on the car and thus find the Coefficient of Friction ( m ) between the car and the ground.


 * Solution:**
 * Velocity is constant so horizontal forces must be equal, **

... .... so Fr = 400 Newtons.


 * Car is not moving up or down, so vertical forces must be equal, **

... ... so N = 900g.

Hence: math \\ . \qquad Fr= \mu N \\. \\ . \qquad 400 = \mu \times 900g \\. \\ . \qquad \mu = \dfrac{400}{900g} \qquad \textit{ Use } g = 9.8 \\. \\ . \qquad \mu \approx 0.045 \quad \textit{(no units)} math

Limiting Friction


 * If a body is stationary but __on the point of moving__, then the same rule for Fr applies.**

As discussed above, when an object on a rough surface is not moving, then the friction acts in the opposite direction to any pulling/pushing force and is equal in magnitude. The object is in **equilibrium**.

If the force is increased then the frictional force increases to match it. This will continue until the object is ** __on the point of moving__ **. At that moment, the frictional force is given by Fr = m N.

If the force continues to increase, the object will begin sliding and the frictional force will remain equal to Fr = m N. {Actually friction drops to a fraction less than the limiting friction value, but we will assume it is the same}

For this reason, **Fr = m N** is called the **Limiting Friction**. It is the upper limit for the value of friction.

If the object is in equilibrium (not moving), the 0 __<__ Fr __<__ m N.

Example 3

A particle of mass 5kg rests on a rough horizontal surface. The coefficient of friction between the particle and the surface is 0.3. Find the greatest horizontal force that can be applied without causing the particle to move.

... ... N = 5g
 * Resolving the forces vertically: **

math \\ . \qquad P = \mu N \\. \\ . \qquad P = 0.3 \times 5g = 1.5g math
 * Resolving the forces horizontally: **

So the max force is 14.7 Newtons. {since g = 9.8}

A 10kg block is sitting on a horizontal surface. A rope attached to the block is being pulled up at 30º above the horizontal with a force of 4 Newton. The block is on the point of moving. Calculate the value of the coefficient of friction.
 * Example 4 **

math \\ . \qquad Fr=4\cos30^\circ \\ .\\ . \qquad Fr=2\sqrt{3} \; \; \text{Newtons} math
 * Resolve the forces horizontally, we can see that: **

math \\ . \qquad N + 4\sin30^\circ = 10g \\. \\ . \qquad N + 2 = 10g \\ .\\ . \qquad N = 10g - 2 math
 * Resolve the forces vertically, **
 * There are two forces acting up: **
 * ** the Normal Reaction force and **
 * ** the vertical component of the pulling force. **
 * Therefore the sum of those two forces must equal the weight force pulling down. **

math \\ . \qquad Fr = \mu \, N \\. \\ . \qquad 2\sqrt{3} = \mu \big( 10g - 2 \big) \\. \\ . \qquad \mu = \dfrac{2\sqrt{3}}{10g-2} \qquad \textit{ Use } g = 9.8 \\. \\ . \qquad \mu = \dfrac{2\sqrt{3}}{96} \\. \\ . \qquad \mu = \dfrac{\sqrt{3}}{48} \\. \\ . \qquad \mu \approx 0.036 math
 * Connecting these two results and using Fr = m N, we get: **

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