AP Physics Section 4-8 Friction

What is friction? Section 4-8 Friction Friction is a force that opposes motion. Friction exists between two surfaces. Friction is related to electrostatic forces between the atoms in the two surfaces.

Classes of friction There are several classes of friction, but two imported classes are: Dry friction — resisting lateral motion between two solid surfaces. Fluid friction — acting between layers in fluids that are moving relative to each other.

What does friction depend on? Dry friction does depend on a component of weight. Dry friction does depend on a component of weight. Dry friction does depend on a component of weight. There is a reaction force exerted by the surface, but only perpendicular to the surface. This is the normal force. So friction is actually proportional to the normal force. There is a reaction force exerted by the surface, but only perpendicular to the surface. This is the normal force. So friction is actually proportional to the normal force. There is a reaction force exerted by the surface, but only perpendicular to the surface. This is the normal force. So friction is actually proportional to the normal force. There is a reaction force exerted by the surface, but only perpendicular to the surface. This is the normal force. So friction is actually proportional to the normal force. There is a reaction force exerted by the surface, but only perpendicular to the surface. This is the normal force. So friction is actually proportional to the normal force. Dry friction does not depend on the surface area. Dry friction does not depend on the surface area. Dry friction does not depend on the surface area. Ff μ (a constant) = FN mu, the coefficient of friction

Two types of dry friction Kinetic friction — acts on moving objects. FK μK FK = μK•FN = FN Static friction — acts on stationary objects. Static friction increases as the force on the object increases. FS (max.) μS FS ≤ μS•FN = FN

Graphing Friction

Free body diagrams It’s important to show all the forces acting on an object. This is done graphically using a free body diagram. Only forces on the object itself are shown, not forces the object exerts on the environment around it. It’s important to show all the forces acting on an object. This is done graphically using a free body diagram. Only forces on the object itself are shown, not forces the object exerts on the environment around it. It’s important to show all the forces acting on an object. This is done graphically using a free body diagram. Only forces on the object itself are shown, not forces the object exerts on the environment around it. It’s important to show all the forces acting on an object. This is done graphically using a free body diagram. Only forces on the object itself are shown, not forces the object exerts on the environment around it. For instance, a free body diagram of a block on a table would show the force of the earth on the block (gravity), and the force of the table on the block, FN. The forces of the block on the table and on the earth would not be shown. For instance, a free body diagram of a block on a table would show the force of the earth on the block (gravity), and the force of the table on the block, FN. The forces of the block on the table and on the earth would not be shown. For instance, a free body diagram of a block on a table would show the force of the earth on the block (gravity), and the force of the table on the block, FN. The forces of the block on the table and on the earth would not be shown. For instance, a free body diagram of a block on a table would show the force of the earth on the block (gravity), and the force of the table on the block, FN. The forces of the block on the table and on the earth would not be shown. FN Fg

Pulling against friction Finding static and kinetic friction when pulling an object over a surface. FN y Start pulling Fstatic (max) Pull enough to begin motion FP Fstatic FP x Fg FS (max) FN = μS

FP FN FK FPy FPx Fg Fg = mg Fg = (27.2kg)(–9.80 m/s2) –266.6 N = Two men pull a 27.2 kg wooden box along a smooth concrete road surface. Once the box is in motion the men pull with a force of 113 N at an angle of 25.0° to the horizontal in order for it to move with constant speed. Find the coefficient of kinetic friction. FP FN FK FPy 25° FPx Fg Fg = mg Fg = (27.2kg)(–9.80 m/s2) –266.6 N =

= = = Horizontal: Fnet(x) = FPx + FK = 0 FPx = FP cosθ = 113 N cos 25° FK = –102.4 N Fnet(y) = Fg + FPy + FN Vertical: = 0 FPy = FP sinθ = 113 N sin 25° FPy = +47.8 N Fg + FPy + FN = 0 FN = –Fg + –FPy FN = –(–266.6 N) + –(+47.8 N) = +218.8 N FK 102.4 μK = = = 0.468 FN 218.8

Motion on an incline Problems with motion on a ramp, or inclined plane, are classic in physics. Problems with motion on a ramp, or inclined plane, are classic in physics. Problems with motion on a ramp, or inclined plane, are classic in physics. Always choose the most convenient coordinate system. On a ramp, this is usually parallel and perpendicular to the surface. Always choose the most convenient coordinate system. On a ramp, this is usually parallel and perpendicular to the surface. Always choose the most convenient coordinate system. On a ramp, this is usually parallel and perpendicular to the surface. Always choose the most convenient coordinate system. On a ramp, this is usually parallel and perpendicular to the surface. Recall that the normal force, FN, is always perpendicular to the surface. It usually balances the perpendicular component of the weight, Fg. Recall that the normal force, FN, is always perpendicular to the surface. It usually balances the perpendicular component of the weight, Fg. Recall that the normal force, FN, is always perpendicular to the surface. It usually balances the perpendicular component of the weight, Fg. Recall that the normal force, FN, is always perpendicular to the surface. It usually balances the perpendicular component of the weight, Fg. Recall that the normal force, FN, is always perpendicular to the surface. It usually balances the perpendicular component of the weight, Fg.

A stationary object on an incline Fnet = Fg + FN + FS = 0 FN Fnet(x) = Fgx + FS = 0 FS FS = –Fgx = –Fg sinθ FS = –mg sinθ Fnet(y) = Fgy + FN = 0 Fgy θ x FN = –Fgy = +Fg cosθ θ FN = +mg cosθ Fg FS mg sinθ Fgx y μs = = = tanθ FN mg cosθ

An accelerating object on an incline Fnet = Fg + FN + FK ≠ 0 FN Fnet(y) = Fgy + FN = 0 a FK FN = –Fgy = +Fg cosθ FN = +mg cosθ Fnet Fnet(x) = Fgx + FK = ma Fgy θ x FK = μK•FN θ Fg (+mg sinθ) + (–μK mg cosθ) = ma Fgx y a = g(sinθ–μK cosθ) a = Fnet /m

Ramp problem θ = 30.0° a = g(sinθ–μK cosθ) μK = 0.18 A sled slides from rest at the top of a snow covered hill. If the hill is inclined at 30.0° and the coefficient of kinetic friction between the sled’s runners and the snow is 0.18, what is the sled’s acceleration? θ = 30.0° a = g(sinθ–μK cosθ) μK = 0.18 a = 9.8 [ (sin 30°) – (0.18 cos 30°) ] a = 9.8 [ (0.50) – (0.18•0.866) ] a = 9.8 [ (0.50) – (0.16) ] a = 9.8 (0.34) a = 3.3 m/s2 down the hill