PHY250L
Lab 4 Friction
Student Name: Click here to enter text.
Kit Code (located on the lid of your lab kit): Click here to enter text.
“Pre-Lab Questions”
1. “If the normal force perpendicular to a surface, what happens to the magnitude of the normal
force on an object as the angle of the incline is increased?”
The Magnitude of the Normal force decreases
as the angle of incline is increased.
2. “Applying Newton’s Second Law and the
equation for static friction (F = μsN)₁ prove that
the coefficient of static friction (μs) is related to
the minimum angle, θ, that causes the block to
slip (Figure 5) by the equation μs = tan(θ).”
Let the mass of the block be "m".
The force due to gravity on the block is "mg".
The component of the force due to gravity along the direction of the slope is mgsinθ.
The component of the force due to gravity perpendicular to the direction of the slope is mgcosθ.
This is in the direction opposite to the normal force on the block, "N", and both these forces cancel each
other so that the block does not move in this direction.
N - mgcosθ = 0
So, N = mgcosθ
For the block to slip, the component of the force due to gravity in the direction of the slope should be
greater than the maximum static friction.
mgsinθ > μsN
To calculate the minimum angle needed for the block to slip,
mgsinθ = μsN
mgsinθ = μsmgcosθ
The ultimate study guide
, PHY250L
Lab 4 Friction
sinθ = μscosθ
Therefore, μs = tanθ
3. “A person applied a horizontal force to a crate of mass, m, that caused the crate to move at a
constant velocity (Figure 6). Show that the
relationship between the applied force and the
normal force is FA = μFN.”
Force due to friction is directly proportional to
the force applied by the frictional surface(FN)
and the constant of proportionality is μ.
So, frictional force = μFN.
When the block moves at a constant velocity,
there is no acceleration on the block, hence,
no net force on it.
The two forces acting on it in the direction of motion are the applied force(FA) and the frictional
force(μFN).
Since there is no net force, these both should cancel out.
FA - μFN = 0
Therefore, FA = μFN
“Experiment 1: Static Friction and Mass on an Inclined Plane”
“Table 1: Wooden Block Incline Data” “Table 2: Metal Washer Incline Data”
“Trial
“Angle”
”
“1” 39
“2” 38
The ultimate study guide
Lab 4 Friction
Student Name: Click here to enter text.
Kit Code (located on the lid of your lab kit): Click here to enter text.
“Pre-Lab Questions”
1. “If the normal force perpendicular to a surface, what happens to the magnitude of the normal
force on an object as the angle of the incline is increased?”
The Magnitude of the Normal force decreases
as the angle of incline is increased.
2. “Applying Newton’s Second Law and the
equation for static friction (F = μsN)₁ prove that
the coefficient of static friction (μs) is related to
the minimum angle, θ, that causes the block to
slip (Figure 5) by the equation μs = tan(θ).”
Let the mass of the block be "m".
The force due to gravity on the block is "mg".
The component of the force due to gravity along the direction of the slope is mgsinθ.
The component of the force due to gravity perpendicular to the direction of the slope is mgcosθ.
This is in the direction opposite to the normal force on the block, "N", and both these forces cancel each
other so that the block does not move in this direction.
N - mgcosθ = 0
So, N = mgcosθ
For the block to slip, the component of the force due to gravity in the direction of the slope should be
greater than the maximum static friction.
mgsinθ > μsN
To calculate the minimum angle needed for the block to slip,
mgsinθ = μsN
mgsinθ = μsmgcosθ
The ultimate study guide
, PHY250L
Lab 4 Friction
sinθ = μscosθ
Therefore, μs = tanθ
3. “A person applied a horizontal force to a crate of mass, m, that caused the crate to move at a
constant velocity (Figure 6). Show that the
relationship between the applied force and the
normal force is FA = μFN.”
Force due to friction is directly proportional to
the force applied by the frictional surface(FN)
and the constant of proportionality is μ.
So, frictional force = μFN.
When the block moves at a constant velocity,
there is no acceleration on the block, hence,
no net force on it.
The two forces acting on it in the direction of motion are the applied force(FA) and the frictional
force(μFN).
Since there is no net force, these both should cancel out.
FA - μFN = 0
Therefore, FA = μFN
“Experiment 1: Static Friction and Mass on an Inclined Plane”
“Table 1: Wooden Block Incline Data” “Table 2: Metal Washer Incline Data”
“Trial
“Angle”
”
“1” 39
“2” 38
The ultimate study guide
