Force, Acceleration and Friction Problems Physics 1 Academic

Problem #1 Rose is sledding down an ice-covered hill inclined at an angle of 15° from horozontal. If Rose and the sled have a combined mass of 54.0 kg, what is the force pulling them down the hill?

The forces on the sled are: Gravitational force (mg) Always acts downward The weight of Rose and the sled From the formula F=mg Normal Force (FN) Acts perpendicular to the surface Pushes against the bottom of the sled The resultant of these forces goes in the direction of motion of the sled Called FD

Here are the forces redrawn as a right triangle Here are the forces redrawn as a right triangle. Θ is the angle of the slope.

Sine The sine of an angle is the opposite side over the hypotenuse. So, in our case the sine Θ = opp./hyp. = FD/mg

Known Unknown Θ = 15° FD=. m = 54 kg g = 10. 0 m/s2 sin 15° = FD/(54 Known Unknown Θ = 15° FD= ? m = 54 kg g = 10.0 m/s2 sin 15° = FD/(54.0 kg)(10.0 m/s2) FD = 140 N

Friction - Definitions Static friction – the resistance force between two surfaces that must be overcome to start an object in motion. Kinetic or sliding friction – the resistance force between two surfaces already in motion. Rolling friction – the resistance force between a surface and a rolling object. Fluid friction – the resistance force of a gas or liquid as an object passes through.

for a horizontal surface FN is the weight in Newtons Friction We will concern ourselves with kinetic or sliding friction. force of sliding friction = coefficient of sliding friction x normal force for a horizontal surface FN is the weight in Newtons

The Magic Triangle FF μ FN

Problem #2 Brian is walking through the school cafeteria but does not realize that the person in front of him has just spilled his milk. As Brian who weighs 420 N, steps in the milk, the coefficient of sliding friction between Brian and the floor is suddenly reduced to 0.040. What is the force of sliding friction between Brian and the slippery floor?

Known Unknown FN = 420 N FF = ? μ= 0.040 FF = (0.040)(420 N) FF = 17 N

Problem #3 While redecoration her apartment, Kitty slowly pushes an 82 kg china cabinet across the wooden dining room floor, which resists the motion with a force of friction of 320 N. What is the coefficient of sliding friction between the china cabinet and the floor?

Known Unknown m = 82 kg w =. g = 10 m/s2 w = mg w = (82 kg)(10 Known Unknown m = 82 kg w = ? g = 10 m/s2 w = mg w = (82 kg)(10.0 m/s2) w = 820 N = FN

Known Unknown FN = 820 N μ = ? Ff = 320 N μ = 0.39

Cosine FN θ = 25° mg

At Sea World, a 900 kg polar bear slides down a wet slide inclined at an angle of 25.0° to the horizontal. The coefficient of friction between the bear and the slide is 0.0500. What is the frictional force that resists the bear’s motion down the slide?

Known Unknown m = 900kg FN = ? g = 10.0 m/s2 θ = 25.0° FN = 8160 N

Known Unknown FN = 8160 N Ff =. μ = 0. 0500 FF = (0 Known Unknown FN = 8160 N Ff = ? μ = 0.0500 FF = (0.0500)(8160N) FF = 408 N