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Lesson 4: Friction
Lesson 4: Friction
CRB True or false? Friction is a type of equilibrium that opposes the movement of objects.
False. Friction is a type of Force that opposes the movement of objects.
CRB Would you be able to consider the Normal Force and Gravitational Force an Action-Reaction pair, as defined by Newton's Third Law?
No. Although the Normal Force and Gravitational Force can be equal and opposite (if not on an inclined plane), they are both acting on the same object, not two objects exerting equal and opposite forces on each other!
Assuming no friction, what would be the net force acting on a box on an incline angled at 30 degrees, with a weight of 10N?
(A) 3
(B) 5
(C) 8
(D) 15
(B) 5
Angle = 30 degrees
*No friction.
*Normal force cancels out the perpendicular force of gravity.
Net force = Parallel force of gravity = 10N x sin30 = 5N
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If a box on an inclined plane is held stationary by friction, and the parallel component to the force of gravity has a force of 10 N, what is the force of friction?
Net force = 0
Force of friction- parallel force of gravity = 0
Force of friction = parallel force of gravity = 10 N
CRB True or false? If you forget to only consider the parallel force of the Gravity when you calculate the Normal Force (and instead consider all Gravity instead), you will underestimate the Frictional Force.
False. If you forget to only consider the parallel force of the Gravity when you calculate the Normal Force (and instead consider all Gravity instead), you will Overestimate the Frictional Force.
Using all of gravity would give you too large of a Normal Force, which directly correlates with friction!
What is the equation for force of static friction in terms of normal force?
0 ≤ static friction ≤ (coefficient of static friction)·(normal force)
Static friction is equal and opposite to the direction of the force along a surface up to the point where it equals the product of the normal force and the coefficient of static friction (μs).
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CRB True or false? The equations for Friction are Vector Equations, so you should use the Right Hand Rule to determine the Friction's Direction.
False. The equations for Friction are NOT Vector Equations. Friction will always be opposing the direction of movement, so there is no need to use the Right-Hand Rule.
CRB Which of the following about the equations for Friction are correct?
I. The equations for Friction are Vector Equations.
II. To determine the direction of the Friction Vector, you must use the Right Hand Rule.
III. The Friction Equations will give you the magnitude of the Friction, but not direction.
(A) I only
(B) II only
(C) III only
(D) I and II only
(C) III only
The Friction Equations are NOT Vector Equations, so they will only give a magnitude of Friction. The direction of Friction will be opposing the direction of motion.
CRB Compare Static Friction and Kinetic Friction, using an MCAT Review Book and a tabletop.
Static friction would apply when the MCAT review book has no velocity, and needs a force to get the book moving on top of the table. Once the book is moving, then Kinetic Friction applies, opposing the book already sliding across the tabletop.
CRB True or false? Car tires under normal driving conditions (rolling along the street) experience Kinetic Friction.
False. Car tires under normal driving conditions (rolling along the street) experience Static Friction.
This is because the tire is not actually sliding across the street, but rather new contact points are being established during the roll. Those contact points will need to have their static friction overcome to keep the tire rolling.
CRB Let's take the same car tire example as last problem, but now the roads are coated in an inch of ice. Trying to take a sharp turn, your tires end up sliding across the ice, and you momentarily lose control. Would this be an example of static or kinetic friction?
Because the tires were actually sliding this time, and not just new contact points being established and broken, this would be an example of Kinetic Friction.
True or false: Even though the calculation of static friction [(coefficient of static friction)·(normal force)]could exceed the other forces acting on an object, static friction will not cause an object to accelerate.
True. Even though the calculation of static friction [(coefficient of static friction)·(normal force)] could exceed the other forces acting on an object, static friction will not cause an object to accelerate.
Remember, this is STATIC friction. At its strongest, it only keeps position the same, not cause acceleration!
A block is sitting on a flat surface, with a weight of 100 N. If it takes a force of 73 N to just barely start moving the block, what is the coefficient of static friction (μs)?
(A) .54
(B) .62
(C) .73
(D) .81
(C) .73
Fs ≤ μs·Fn
73 = μs x 100
μs = 73 N/100 N
μs = 0.73
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What is the equation for kinetic friction in terms of normal force?
kinetic friction = (coefficient of kinetic friction)·(normal force)
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CRB True or false? Unlike Static Friction, which could vary with the amount of force being applied, Kinetic Friction is a constant value for any given coefficient of friction and normal force, meaning that it can cause deceleration.
True. Unlike Static Friction, which could vary with the amount of force being applied, Kinetic Friction is a constant value for any given coefficient of friction and normal force, meaning that it can cause deceleration.
What is the relationship between the coefficient of static friction and the coefficient of kinetic friction?
coefficient of kinetic friction ≤ coefficient of static friction
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CRB Suppose that two football teammates are pushing the practice sled. After a few seconds, one of the teammates stops pushing, but the other person can still move the sled. That second player then stops, catches his breath, and tries to move the sled again by himself. This time, it won't budge. How would you explain this phenomenon using friction?
Based on this description, the second football player could provide enough force to overcome the kinetic friction, but not enough to overcome the static friction. Thus, he could move the sled once it was already moving, but could not start moving the sled from a stand-still.
A box with a mass of 1 kg is pushed by a force of 20N. The coefficients of static and kinetic friction are .5 and .4, respectively. What is the net force on the box the moment it starts moving (in N)?
(A) 16.1
(B) 23.3
(C) 15.1
(D) 34.2
(C) 15.1
Mass = 1 kg
Coefficient of static friction = .5
Normal force = (perpendicular force)
Normal force = gravity = 1 kg x 9.8m/s^2 = 9.8 N
Friction = normal force x coefficient of friction
Static friction = 9.8 N x .5 = 4.9 N
Net force = Sum of all forces
Gravity and Normal force cancel out, force - friction
Net force at beginning of movement = 20 N - 4.9 N = 15.1 N
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A box with a mass of 1 kg is pushed by a force of 20N. The coefficients of static and kinetic friction are .5 and .4, respectively. What is the net force on the box while it is moving (in N)?
(A) 16.1
(B) 23.3
(C) 15.1
(D) 34.2
(A) 16.1
Mass = 1 kg
Coefficient of kinetic friction = .4
Normal force = (perpendicular force)
Normal force = gravity = 1 kg x 9.8m/s^2 = 9.8 N
Friction = normal force x coefficient of friction
Net force = Sum of all forces
Gravity and Normal force cancel out, force - friction
Kinetic friction = 9.8 N x .4 = 3.9 N
Net force while moving = 20 N - 3.9 N = 16.1 N
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CRB Another example where the Normal Force is not necessarily equal to Gravity is when there is a vertical net force. For example, when riding in an elevator down 10 floors, the elevator slows down before stopping. While the elevator is slowing its descent, would the Normal Force be greater than or less than the force of Gravity?
(Hint: Draw a Free Body Diagram!)
Because the elevator is accelerating upwards, the Net Force must be going up. This means that the Normal Force must be greater than the force of Gravity!
