Have you ever been sledding only to sit down on your sled and not be able to get it moving? Did you start scooting or try to make your sled slicker so that you could sled? Did you pack down the snow some more so that the sled with slide quickly? In essence, you are trying to decrease the friction between the sled and the snow. What changes the friction between two surfaces depends on two things: the surfaces in contact with each other and the amount of force pushing the two surfaces together. It is always in a direction that opposes the motion of the object. This can be summarized using an equation that was introduced by an Australian physicist by the name of Augustine de Coulomb in 1794.
Force of Friction
\(\large\mathsf{ F_{friction} = \mu F_{normal} }\)
...where \(\small\mathsf{ \mu }\) is the coefficient of friction between two surfaces.
The coefficient of friction is found through experimental procedures and it depends on whether the two surfaces are not moving relative to each other or are moving relative to each other. Click through the slideshow below to see how the value of \(\small\mathsf{ \mu }\) is found experimentally.
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Set up of Experiment A known mass made of a specific substance is set on an incline of a specific type of surface. The mass will want to slide, but it is kept from moving by the force of friction between it and the surface. Determining Limiting Friction The incline of the ramp is increased until the frictional force can no longer keep the mass from sliding. Right before it starts sliding is the maximum force of friction called the limiting friction. The normal force balances the component of the weight force that is perpendicular to the surface of the incline. The limiting friction balances the component of the weight force that is parallel to the surface of the incline. The coefficient of friction then becomes the ratio between the magnitude of the normal force and the magnitude of the force of friction. \(\mathsf{ \mu = \frac{F_{friction}}{F_{normal}} }\) Since the vales of the forces are found experimentally, the values for the coefficient of friction are tabulated and recorded and range anywhere from 0.01 to 1 depending on the surfaces involved. Since this takes into consideration the amount of force necessary to START the object moving, this is considered the static situation or \(\mathsf{ \mu_s }\). Determining Coefficient of Kinetic Friction The amount of force needed to start a object in motion is more than the force that is needed to keep it in motion. That's because if the two surfaces are actually moving, they do not have time to create as many atomic bonds to create the frictional force. To find the force needed to oppose the kinetic friction, you place the object on a level surface (horizontal). The normal force is equal to the weight force in this situation and the amount of applied force necessary to keep the object moving at a constant speed must equal the force of friction. \(\mathsf{ \mu = \frac{F_{friction}}{F_{normal}} }\) Since this takes into consideration the amount of force necessary to KEEP the object moving, this is considered the kinetic situation or \(\mathsf{ \mu_k }\). Again, these forces are found experimentally specific to the two surfaces involved, so like the coefficients of static friction, the coefficients of kinetic friction are recorded. The kinetic coefficients are smaller than the static coefficients. Determining Which Coefficient to Use In problem solving, take note of whether an object is moving or not. That will determine which value for the coefficient of friction you use in your problem solving. Regardless, the equation stays the same, but the subscript of the coefficient will change with the situation. Object is NOT Moving \(\mathsf{ F_{friction} = \mu_s F_{normal} }\) Object IS Moving \(\mathsf{ F_{friction} = \mu_k F_{normal} }\) |
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Question
What is the force of static friction between a 10.0 kg mass and a surface if the coefficient of static friction is 0.20?
\(\mathsf{ F_{friction} = \mu_s F_{normal} }\)
\(\mathsf{ F_{friction} = (0.20)(10.0 \text{ kg})(9.81 \text{ m/s}^2) }\)
\(\mathsf{ F_{friction} = 19.6 \text{ N} }\)
The direction of this force would be opposing the direction of motion or opposing the applied force.
