Friction and pulleys solved problems
Calculating friction and pulley problems notes
▲
According to the second law of motion we have
F = m a
In the gravitational system the force acting on a mass is
W = m g
[
kgm m / s
^{2}
]
A force of 1 Newton is equal to
N = 1
[
kgm m / s
^{2}
]
Then if the
weight
is given it equals
W
[
kgf
]
= 9.8 N
There are two friction coefficients the static μ
_{s}
and the the dynamic or kinetic μ
_{k}
The static friction coefficient value is greater then the kinetic coefficient value ( μ
_{s}
> μ
_{k}
), it means that we have to employ greater force in order to start the movement of the mass, but once the mass is moving we have to employ less force to keep it moving.
In many problems we assume that both coefficients are the same, if not we have to relate to the proper coefficient e.g. if the bodies are moving we have to take the kinetic friction coefficient.
In many questions we have to decide the direction of the movment, it is important because the direction of the friction force is always opposite to the movment direction.
If we look at the figure at left the direction of the movment is to the right as the direction of a
_{1}
the equilibrium equation will be:
ΣF
_{x}
= F − m g sin θ − f = m a
and the acceleration is:
If the movment is downward as in the second figure at left then the equilibrium equation will be:
ΣF
_{x}
= m g sin θ − F − f = m a
and the acceleration is:
We can clearly see the difference of the signs and the values of the accelerations in both direction of the chose directions.
The conclusion frome this calculation is that in friction problems a negative acceleration result doesn't mean that the acceleration is in the opposite direction but rather we have to change the direction of the acceleration and solve again, if this time we got negative value then the mass will not move due to the friction force, for more information see
example 7
,
example 38
.
If mass M is moving a distance of x meter then this distance is divided by each cable that holds mass m and mass m is moving a distance of x/2 meter.
The
distance traveled
is given by:
x = v
_{0}
t + a t
^{2}
/ 2
At rest v
_{0}
= 0 and the acceleration is:
a = 2 x / t
^{2}
time t of both masses M and m is the same so we can see that the acceleration is related to the distance, if the distance is half also the acceleration will be half so:
a
_{1}
= 2 a
_{2}
For system rest we need that the acceleration will be 0. From the equation
v
_{t}
= v
_{0}
+ a t
we get the result
v
_{t}
= v
_{0}
Friction example - 1
▲
A body whose mass is m is resting on an inclined surface with an angle of θ and a friction coefficient of μ, find the acceleration of the mass m as a function of μ and angle θ.
From equilibrium in the x and y direction we get:
ΣF
_{x}
= m g sinθ − N μ = m a
(1)
ΣF
_{y}
= N − m g cosθ = 0
(2)
From eq.
(2)
N = m g cosθ
Substitute N into eq.
(1)
m g sinθ − m g μ cosθ = m a
And the acceleration is:
a = g( sin θ − μ cos θ)
When the system is at rest a = 0 then:
We can see that the friction coefficient for rest condition must be equal or greater then the tangent of the slope μ > tan θ.
Friction example - 2
▲
A body whose mass is m is resting on a horizontal surface whose friction coefficient is μ, a force F is implied on the mass at an angle of θ degree. Find the acceleration of mass m and the maximum force that the mass m will still be at rest.
From the forces diagram we can derive the equilibrium equations:
ΣF
_{x}
= F cos θ − Nμ = ma
(1)
ΣF
_{y}
= N + F sin θ − mg = 0
(2)
From eq.
(2)
N = mg − F sin θ
Substitute N into eq.
(1)
F cos θ − mgμ + Fμ sin θ = ma
Solving for a we get:
(3)
For the maximum force F that can be applied so that the mass m will still be at rest we have:
f
_{max}
= N μ
_{s}
= (m g − F sinθ) μ
_{s}
(4)
(5)
A force of 30 N is applied to a mass of 3 kg for 5 sec at an angle of 30 degree, if the static and kinetic frictions are 0.5 and 0.2 respectively. Find the acceleration and the time needed for mass m to stop after force F is eliminated.
First check if the mass will move by the condition: f
_{max}
<
F cosθ.
Because 7.2
<
25.98 the mass will move and the acceleration is according to equation (3).
The velocity of the mass after 5 sec is found from the
acceleration aquation
:
v
_{t}
= v
_{0}
+ a t = 0 + 7.7 * 5 = 38.5 m/s
The new free body diagram after eliminating force F is at left.
According to the second law of motion F = m a (F now is only the friction force) and f = − m a (the minus sign is because of the opposite directions of a and f).
a = − f / m = − m g μ
_{k}
/ m = −g μ
_{k}
= − 9.8 * 0.2 = − 1.96 m/s
^{2}
And the time untill mass m will stop is according to the equation: v
_{t}
= v
_{0}
+ a t where v
_{t}
= 0 and v
_{0}
= 38.5 m/s
t = − v
_{0}
/ a = − 38.5 / − 1.96 = 19.64 sec
Friction of two masses - example 4
▲
Two masses are located on each other, mass m is tied to the wall by a rope, a force F is exerted on mass M , find the ecceleration of mass M if the friction coefficients between the two masses is μ
_{1}
and μ
_{2}
between mass M and the lower surface.
Friction force f
_{1}
is:
f
_{1}
= mgμ
_{1}
Friction force f
_{2}
is:
f
_{2}
= (M + m)gμ
_{2}
Mass M is effected by both frictions f
_{1}
and f
_{2}
and is:
f
_{M}
= f
_{1}
+ f
_{2}
= mgμ
_{1}
+ (M + m)gμ
_{2}
From the forces diagram we can derive the equilibrium equations in the x direction on both masses:
On mass m:
ΣF
_{x}
= T
_{1}
− mgμ
_{1}
= 0
(1)
On mass M:
ΣF
_{x}
= F − f
_{1}
− f
_{2}
= Ma
(2)
Maximum force F that can be applied to mass M without sliding is:
F <= mgμ
_{1}
+ (M + m)gμ
_{2}
From eq (1) T
_{1}
can be found:
T
_{1}
= mgμ
_{1}
Two masses and pulley - example 6
▲
Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.
The kinetic friction force is equal to:
f = μ
_{k}
N = μ
_{k}
M g
From the free body diagram on mass m and M and assuming that the acceleration is downward we get:
On mass M:
ΣF
_{x}
= T − M g μ
_{k}
= M a
(1)
On mass m:
ΣF
_{y}
= m g − T = m a
(2)
Eliminating T from both equations gives:
m g − M g μ
_{k}
+ M a = −m a
(3)
Mass M will accelerate when
m > μ
_{s}
M
System of the masses are at rest when
m <= μ
_{s}
M (then a = 0)
Find the friction coefficient and the tension in the rope if M = 5 kg and m = 1 kg the masses are moving at a constant speed.
Because the masses are moving at a constant speed it is necessary that a = 0. from eq. (3) we have
m − M μ
_{k}
= 0
The tension in the rope can be found by solving eq. (1) and (2) :
Two masses and pulley - example 7
▲
Figure 1
Figure 2
M
kg
m
kg
θ
deg
μ
a
m/s
_{2}
T
N
f
N
Find the acceleration of the system of masses, the masses of the rope and the pulleys are negligible, assume first that the acceleration of M is upward and second time that the acceleration is downward.
The friction force is equal to:
f = μ N = μ M g cosθ
From the forces diagram on mass m and M and assuming that the acceleration is upward (Figure 1) we get:
On mass M:
ΣF
_{x}
= T − M g sinθ − f = M a
(1)
On mass m:
ΣF
_{y}
= m g − T = m a
(2)
(3)
Solving the case with the acceleration downward (Figure 2).
On mass M:
ΣF
_{x}
= M g sinθ − T − f = M a
(4)
On mass m:
ΣF
_{y}
= T − m g = m a
(5)
(6)
Check the case when M = 2 kg m = 1.2 kg angle θ = 30 degree and the friction coefficient is 0.15
Because both accelerations in both directions are negative motion is not possible in this conditions of masses, friction and slope angle (see note 2).
Determine the acceleration and the motion direction if M = 4 kg
m = 2 kg angle θ = 45 degree and the friction coefficient is 0.1
First check the case of upward acceleration (Figure 1).
Because a is negative and the present of friction we have to make the calculation again assumming this time that the acceleration is downward see the case of (Figure 2).
Notice that we got a different value for the acceleration, the conclution is that mass M is slidding downward.
Note 1:
from eq. (3) and (6) and comparing the acceleration to zero we can derive the range of mass m
that the system will stay at rest.
M(sinθ − μ cosθ) < m < M(sinθ + μ cosθ)
Note 2:
to determine the direction of the motion we can solve the equilibrium equation by eliminating the friction, once we have the direction of the motion we can set the correct direction of the friction force (always opposite to the motion direction) and solve the equilibrium equations. We have to remember that if we found motion in any direction without friction, the friction force can reduce the motion acceleration or even stope the motion.
Two masses on inclined surfaces example - 9
▲
Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulley, the frictions coefficients and the slopes are given, also find the tension in the rope.
To determine the possible direction of the motion we will first solve the forces equations by neglecting the frictions, the results are the conditions:
m sinβ > M sinα
motion can be to the right
(a)
m sinβ < M sinα
motion can be to the left
(b)
m sinβ = M sinα
motion not possible
(c)
Those conditions alone are not enough to verify if the system will move, in order to make the system to slipe the masses have to overcome the friction forces, so the conditions for motions are.
m ( sinβ − μ
_{2s}
cosβ ) > M ( sinα + μ
_{1s}
cosα )
motion is to the right
(d)
M ( sinα − μ
_{1s}
cosα ) > m ( sinβ − μ
_{2s}
cosβ )
motion is to the left
(e)
From the forces diagram and assuming that the acceleration is to the right we get:
On mass M:
ΣF
_{x}
= T − M g sinα − M g cosα μ
_{1k}
= M a
(1)
On mass m:
ΣF
_{x}
= m g sinβ − T − m g cosβ μ
_{2k}
= m a
(2)
System is at rest when:
m(sinβ − μ
_{2s}
cosβ) − M(sinα + μ
_{1s}
cosα) = 0
In order to get positive tension in the rope we need that at least one of the equations will be true:
μ
_{1s}
< tan
α
or
μ
_{2s}
< tan
β
Given two masses M = 40 kg and m = 20 kg, connected with an ideal pulley, the masses are located on two surfaces whose friction coefficients are μ
_{1s}
= μ
_{1k}
= 0.15 and μ
_{2s}
= μ
_{2k}
= 0.25, the slopes has angles of α = 30 deg and β = 53 deg. Find the direction and the value of the acceleration and the tension in the rope which connects both masses.
M
kg
α
deg
m
kg
β
deg
μ
_{1s}
μ
_{1k}
μ
_{2s}
μ
_{2k}
a
m/s
^{2}
T
N
No friction move:
Each mass move:
M
m
System move:
Line
Possibilities
Description
No friction move
←
→
Checks the direction of motion in the case that
the friction is eliminated from the system.
Each mass move
M
m
Checks if motion occurs for each mass separately in the case that the rope connecting both masses is eliminated and only friction forces are present μ < tanα
* Green M is to mark that M will slide.
* Red m is to mark that m will not slide.
System move
Slide left
Slide right
No motion
Describes if the system of masses will slide to the left or to the right or it will not slide at all.
Notes:
1.
When system moves to the right or left the calculation of the tension is straightforward according to the equations found before.
2.
When the system has no motion the maximum tension can be from 0 to the minimum static friction force calculated separately for each mass (see note 3).
3.
In the case of no motion we have to analyze the conditions of the 'Each mass move'. if both masses are red then the tension can be artificially set from 0 to the minimum friction value between both masses (not the maximum value because this will cause the mass of minimum value to move and reduce the tension). If one mass is red and the other is green then the tension is the value of the green mass friction (minimum friction force).
First determine the possible motion direction according to creteria (a) and (b)
m sinβ = 20 sin53 = 16
<
M sinα = 40 sin30 = 20
The direction of the acceleration will be to the left.
Now check creterion (c) and (d) to verify is motion occures:
Motion equations for the case with acceleration to the left are:
On mass M:
ΣF
_{x}
= M g sinα − T − M g cosα μ
_{1k}
= M a
(3)
On mass m:
ΣF
_{x}
= T − m g sinβ − m g cosβ μ
_{2k}
= m a
(4)
Because we already found that the motion is to the left we have to check condition (d) only and we get 14.8 > 12.9 so the masses will slip to the left and the acceleration is:
Friction of two masses and pulley - example 11
▲
Find the acceleration and the tension in the rope of the system of masses shown
(
M
>
m
)
, neglecting the mass of the string and the inertia of the pulley assume that the static and kinetic friction coefficients are equal to μ
_{1}
μ
_{2}
and the slope angle θ are known.
Notice that the two masses are moving in the oposite directions so the friction forces are:.
Friction forces on mass M are:
f
_{1}
= (M + m)gμ
_{1}
cosθ
Friction forces on mass m are:
f
_{2}
= mgμ
_{2}
cosθ
From the forces diagram on mass m and M and assuming that the acceleration of mass M is downward and is equal to the acceleration of mass m in the upward direction we get:
On mass m
ΣF
_{x}
= T − f
_{2}
− mg sinθ = ma
(1)
On mass M
ΣF
_{x}
= Mg sinθ − f
_{1}
− f
_{2}
− T = Ma
(2)
Subsitute the values of f
_{1}
and f
_{2}
into eq. 1 and 2 and eliminating T from both equations we get a:
Two masses on pulley - example 12
▲
Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulley.
From the forces diagram on mass m and M and assuming that the acceleration is to the downward direction of mass M we get:
On mass M:
ΣF
_{y}
= Mg − T = Ma
(1)
On mass m:
ΣF
_{y}
= T − mg = ma
(2)
Solving for the acceleration we get:
And the tensions in the cables are:
Friction of two masses and pulley - example 13
▲
Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulley, the friction coefficient between mass M and the surface is μ.
From the forces diagram on mass m and M and assuming that the acceleration is to the downward direction and notice that the acceleration of mass M is twice the accelaretion of mass m because when mass M is moving a length of x mass m will travel only half this distance.
Forces on mass M:
ΣF
_{x}
= T − Mgμ = Ma
_{1}
(1)
Forces on mass m:
ΣF
_{y}
= mg − T
_{1}
= ma
_{2}
(2)
From acceleration:
a
_{1}
= 2a
_{2}
(3)
From the pulley:
T
_{1}
= 2T
(4)
We got four equations with four unkowns T T
_{1}
a
_{1}
and a
_{2}
, solving the equations we get:
The accelerations are:
The tensions are:
Two masses and two pulleys - example 14
▲
Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulleys.
Forces on pulley 1:
ΣF
_{y}
= 2T − Mg = Ma
_{1}
(1)
Forces on mass m:
ΣF
_{y}
= mg − T = ma
_{2}
(2)
Accelerations balance:
a
_{2}
= 2a
_{1}
(3)
The relation between the accelerations can be found by measuring the distances that the masses travels, while mass m is moving a distance of x mass M will move a distance of x/2 so the acceleration of M is half the acceleration of m.
Solving for a
_{1}
a
_{2}
and T we get:
And the tension in the rope is:
The tension T is uniform along the cable.
The value of T
_{1}
is:
T
_{1}
= 2T
Notice that if we apply a force F downward at mass m we could lift a mass of 2F, so this system multiplies the force by 2.
Relative motion of masses - example 16
▲
m
_{1}
kg
m
_{2}
kg
m
_{3}
deg
a
_{1}
m/s
^{2}
a
_{2}
m/s
^{2}
a
_{3}
m/s
^{2}
T
_{1}
N
T
_{2}
N
Acceleration directions
m
_{1}
m
_{2}
m
_{3}
Find the acceleration of the system of masses neglecting the mass of the string and the mass of the pulley.
We assume arbitrary that all three accelerations are downward and are equal to a
_{1}
a
_{2}
and a
_{3}
. Now we can write the forces acting on each mass as:
ΣF
_{y}
on mass m
_{1}
m
_{1}
g − T
_{1}
= m
_{1}
a
_{1}
(1)
ΣF
_{y}
on mass m
_{2}
m
_{2}
g − T
_{2}
= m
_{2}
a
_{2}
(2)
ΣF
_{y}
on mass m
_{3}
:
m
_{3}
g − T
_{2}
= m
_{3}
a
_{3}
(3)
ΣF
_{y}
on pulley B:
T
_{1}
= 2 T
_{2}
(4)
We can see that we have 5 unknowns and only 4 equations so we have to derive another equation from the accelerations, a
_{1}
is equal to the acceleration of pulley B which sould be half the accelerations of a
_{2}
and a
_{3}
but in negative sign because when a
_{1}
is going down pulley B is going up, but we chose the accelerations in the downward direction.
Σ Accelerations:
a
_{1}
= −
(
a
_{2}
+ a
_{3}
) / 2
(5)
a
_{3}
= − 2a
_{1}
− a
_{2}
The minus sign is because a
_{1}
is in the opposite direction to a
_{2}
and a
_{3}
. Substitute the value of T
_{1}
from eq. (4) and a
_{3}
from eq. (5) to eq. (1) (2) and (3) to get the matrix form:
Solving by
Cramer's rule
we have:
D = 2 (m
_{2}
m
_{3}
+ m
_{2}
m
_{3}
) + m
_{1}
(m
_{3}
+ m
_{2}
) = m
_{1}
m
_{2}
+ m
_{1}
m
_{3}
+ 4m
_{2}
m
_{3}
After division of the numerator and the denominator by m
_{1}
m
_{2}
m
_{3}
we got another form for T
_{2}
If all masses are the same and equal to M then:
T
_{2}
= 2 g M / 3
Friction of two masses example - 17
▲
from the forces diagram we can write the equilibrium equations:
On mass m
_{1}
ΣF
_{x}
= T
_{1}
= f
_{1}
= m
_{1}
g μ
_{1}
(1)
On mass m
_{2}
ΣF
_{x}
= T
_{2}
− f
_{1}
− f
_{2}
= m
_{2}
a
T
_{2}
− m
_{1}
g μ
_{1}
− (m
_{1}
+ m
_{2}
) μ
_{2}
g = m
_{2}
a
(2)
On mass m
_{3}
ΣF
_{y}
= m
_{3}
g − T
_{2}
= m
_{3}
a
(3)
(4)
From eq.
(3)
T
_{2}
can be found
T
_{2}
=
m
_{3}
(
g − a
)
=
Q:
Given three masses m
_{1}
equal to m, mass m
_{2}
equal to 3m and m
_{3}
equal to 2m, μ
_{1}
= 0 and masses m
_{2}
and m
_{3}
are moving at a constant velocity. Find the value of μ
_{2}
, T
_{1}
and T
_{2}
S:
For steady velocity we need that a = 0
From the acceleration equation
(4)
we have:
m
_{3}
− m
_{1}
μ
_{2}
− m
_{2}
μ
_{2}
=
0
Because there is no friction between surfaces of masses m
_{1}
and m
_{2}
then: T
_{1}
= 0
Because the acceleration is 0 the tension T
_{2}
in cable m
_{3}
is: T
_{2}
= m
_{3}
g
Friction of two masses example - 21
▲
Three masses are hanging on frictionless pulleys as shown in the figure. Find the acceleration of the masses.
We choose arbitrary the direction of the accelerations of masses m
_{1}
and mass m
_{2}
as upward and the acceleration of m
_{2}
downward.
From the forces diagram on the masses we have.
On mass m
_{1}
ΣF
_{y}
= T − m
_{1}
g = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= m
_{2}
g − 2T = m
_{2}
a
_{2}
(2)
On mass m
_{3}
:
ΣF
_{y}
= T − m
_{3}
g = m
_{3}
a
_{3}
(3)
From accelerations:
(4)
Note: when mass m
_{1}
is moving x
_{1}
distance upward and mass m
_{3}
a distance of x
_{3}
upward, the contribution to the displacement of mass m
_{2}
is the sum of half of this displacements x
_{2}
= (x
_{1}
+ x
_{3}
) /
2
in the downward direction.
The equations with the unknowns T, a
_{1}
, a
_{2}
and a
_{3}
can be solved by
Cramer's rule
or by direct substitution.
Write the equations in the matrix form
D = − m
_{2}
m
_{3}
− 4m
_{1}
m
_{3}
− m
_{1}
m
_{2}
= − (m
_{1}
m
_{2}
+ m
_{2}
m
_{3}
+ 4m
_{1}
m
_{3}
)
Find the acceletations and the tension in the rope if all the masses are equal to m
_{1}
= m
_{2}
= m
_{3}
= m
T = 2 m g / 3
a
_{1}
= − g / 3
a
_{2}
= − g / 3
a
_{3}
= − g / 3
T = 2 m / 3
Notice
that if m
_{1}
= m
_{3}
= m and m
_{2}
= 2m then the accelerations will be 0 and T = m
If the result of the acceleration is negative then the motion is opposite to that chosen.
Friction of two masses example - 24
▲
Three masses are connected with a rope of 0 mass as shown in the figure, a force F is apply to mass m
_{3}
, friction coefficient μ if present between mass m
_{1}
and the surface that is tilted by θ deg. Find the value of the force F that will keep the system at rest, if the force F is cancelled find the acceleration of the system.
In order to determine the direction of the motion we have to analyze the relationship of the equations:
m
_{3}
>
m
_{2}
+ m
_{1}
sinθ
Motion is m
_{3}
down
(a)
m
_{3}
<
m
_{2}
+ m
_{1}
sinθ
Motion is m
_{3}
up
(b)
m
_{3}
=
m
_{2}
+ m
_{1}
sinθ
No motion
(c)
Assume that the acceleration of mass m
_{3}
is downward then:
On mass m
_{1}
ΣF
_{x}
= T
_{1}
− μ m
_{1}
g cos θ − m
_{1}
g sinθ = m
_{1}
a
(1)
On mass m
_{2}
ΣF
_{y}
= T
_{2}
− m
_{2}
g − T
_{1}
= m
_{2}
a
(2)
On mass m
_{3}
ΣF
_{y}
= m
_{3}
g + F − T
_{2}
= m
_{3}
a
(3)
The same equations when the acceleration is to the left are:
On mass m
_{1}
ΣF
_{x}
= m
_{1}
g sinθ − T
_{1}
− μ m
_{1}
g cosθ = m
_{1}
a
(4)
On mass m
_{2}
ΣF
_{y}
= m
_{2}
g + T
_{1}
− T
_{2}
= m
_{2}
a
(5)
On mass m
_{3}
ΣF
_{y}
= T
_{2}
− m
_{3}
g − F = m
_{3}
a
(6)
We got three equations with three unknowns T
_{1}
T
_{2}
and a.
Solving eq.
(1) (2)
and
(3)
according to
Cramer's rule
we get:
Solving eq.
(4) (5)
and
(6)
according to
Cramer's rule
we get:
The minimum force F nedded to apply when a = 0, when the motion is intended to be to the left is:
F = g ( m
_{1}
sinθ + m
_{2}
− m
_{3}
) − m
_{1}
μ
_{s}
g cosθ
The maximum force F nedded to apply when a = 0, when the motion is intended to be to the right is:
F = g ( m
_{1}
sinθ + m
_{2}
− m
_{3}
) + m
_{1}
μ
_{s}
g cosθ
In the case when F = 0 mass m
_{3}
that keeps the system at rest is in the range of:
m
_{1}
(
sinθ − μ cosθ
) +
m
_{2}
<
m
_{3}
<
m
_{1}
(
sinθ + μ cosθ
) +
m
_{2}
If all three masses are of the same weight M then the force F at rest should be in the range:
M g
(
sinθ − μ
_{s}
cosθ
)
<
F
<
M g
(
sinθ
+
μ
_{s}
cosθ
)
Friction of two masses example - 33
▲
Three masses are connected to each other by ropes of 0 mass, a force F is apply to mass m
_{3}
. Find the value of the tension in the ropes and the acceleration of the masses if a friction coefficient of μ exists between the masses and the surface.
From the free body diagram the friction forces and the equilibrium equations of the masses are:
Friction forces:
f
_{1}
= m
_{1}
g μ
f
_{2}
= m
_{2}
g μ
f
_{3}
= m
_{3}
g μ
ΣF
_{x}
= T
_{1}
− m
_{1}
g μ = m
_{1}
a
(1)
ΣF
_{x}
= T
_{2}
− T
_{1}
− m
_{2}
g μ = m
_{2}
a
(2)
ΣF
_{x}
= F − T
_{2}
− m
_{3}
g μ = m
_{3}
a
(3)
We got three equations with three unknowns T
_{1}
T
_{2}
and a
According to
Cramer's rule
the value of the coefficients determinant is
D = − (m
_{1}
+ m
_{2}
+ m
_{3}
)
Another way to solve the problem is to look on the three masses as one mass equal to M = m
_{1}
+ m
_{2}
+ m
_{3}
From the free body diagram we have:
F − M g μ = M a
and a is:
Once we found the acceleration a we can use eq. (1) and (2) to find T
_{1}
and T
_{2}
Friction of two masses example - 37
▲
Two masses m
_{1}
and m
_{2}
are located on the floor the masses then conected by a pulley as shown in the figure. Find the acceleration of the masses and the tension in the rope connecting the masses if a force of F is apply to the pulley upward, suppose that m
_{1}
> m
_{2}
.
We mark the accelerations of the masses as a
_{1}
and a
_{2}
, from the free body diagram we get:
On mass m
_{1}
:
ΣF
_{y}
= T − m
_{1}
g = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= T − m
_{2}
g = m
_{2}
a
_{2}
(2)
On the pulley:
ΣF
_{y}
= F = 2 T
(3)
From eq.
(3)
we get:
T = F / 2
And substituting T to eq.
(1)
and
(2)
we get:
Notes:
because accelerations a
_{1}
and a
_{2}
can not be negative (opposite to the direction shown) because of the floor there are some unique casses:
Samll mass m
_{2}
F < 2 m
_{2}
g
The force F is not enough to lift any one of the masses
Bigger mass m
_{1}
2 m
_{2}
g < F < 2 m
_{1}
g
The force F will lift the small mass but not the bigger mass
2 m
_{1}
g < F
The force F will lift both masses
For example if mass m
_{1}
= 30 kg and m
_{2}
= 20 kg then find the accelerations when F = 100N, F = 400N and F = 1000N.
F
[N]
a
_{1}
[m/s
^{2}
]
a
_{2}
[m/s
^{2}
]
Notes
100N
100 / 30 − 9.8 = − 6.5
100 / 20 − 9.8 = − 4.8
No lift of any mass a
_{1}
= 0 a
_{2}
= 0
200N
200 / 30 − 9.8 = − 3.1
200 / 20 − 9.8 = 0.2
Only small mass is lifted a
_{1}
= 0
1000N
400 / 30 − 9.8 = 3.5
400 / 20 − 9.8 = 10.2
Both masses moves upward
Friction of two masses a force and pulley - example 38
▲
m
_{1}
kg
m
_{2}
kg
μ
F
N
θ
deg
a
m/s
_{2}
T
N
f
N
Mass m
_{1}
and m
_{2}
are at rest when a force F is applied at an angle of θ degree. Find the maximum force F that the system will stay at rest and the acceleration of the system.
The friction force acting on m
_{1}
is:
f = N μ = (m
_{1}
g + F sin θ) μ
.
In order to determine the possible direction of the motion we have to check the forces acting on mass m
_{1}
without friction force.
if
F cosθ
>
m
_{2}
g
m
_{1}
moves to the left
(a)
if
F cosθ
<
m
_{2}
g
m
_{1}
moves to the right
(b)
After we determined the direction of the motion we can set the right direction of the friction force and check if there is motion at all due to the friction force.
Let assume that the acceleration of m
_{1}
is to the left, from the free body diagram we have
On m
_{1}
F cosθ − f − T = m
_{1}
a
(1)
On m
_{2}
T − m
_{2}
g = m
_{2}
a
(2)
From eq. (1) and (2) we can find the values of a and T
If we asumme that the acceleration of mass m
_{1}
is to the right direction the equilibrium equations will be as follows:
On m
_{1}
T − F cosθ − f = m
_{1}
a
(3)
On m
_{2}
m
_{2}
g − T = m
_{2}
a
(4)
If we get a negative value for the acceleration in both directions then motion is not possible.
The conditions for motion are:
If
F cosθ − f
>
m
_{2}
g
m
_{1}
moves to the left
(c)
If
F cosθ + f
<
m
_{2}
g
m
_{1}
moves to the right
(d)
The range of the force F that the system is at rest is when it is equal to the friction force in both directions:
the static and dynamic friction coefficient between the surface and m
_{1}
is μ = 0.4. Find the friction force acting on m
_{1}
if the force equal 250N and θ = 30 degree and m
_{1}
= 10kg and m
_{2}
= 30 kg.
First we check conditions (c) and (d) to verify if the masses are moving.
Condition
(c)
30 * g > 250 * cos30 + (10 * g + 250 * sin30) 0.4
False
Condition
(d)
30 * g < 250 * cos30 - (10 * g + 250 * sin30) 0.4
False
Because both conditions are false the system is at rest.
a) Friction force can not be calculated by the equation for f because there is no movement, f will be calculated by the difference between force of mass m
_{2}
g and F * cosθ
f = m
_{2}
g − F * cosθ
= 30 * 9.8 − 250 * cos30 = 77.5 N
b) maximum force F
F = g(10*0.4 + 30) /(cos 30 - sin 30 * 0.4) = 500.3 N
c) acceleration when F = 0
a = −g(10*0.4 - 30) /(10 + 30) = 6.4 m / s
^{2}
Two masses and two pulleys example - 40
▲
Two masses m
_{1}
and m
_{2}
are connected through 2 pulleys as shown in the figure the friction of m
_{1}
and the surface is μ. Find the accelerations and the tensions in the ropes.
In order to find the direction of the motion we will analyze the forces on the pulley by neglecting the friction force, it is easy to see that if:
m
_{1}
sinθ
>
2 m
_{2}
m
_{1}
is moving downward
(a)
m
_{1}
sinθ
<
2 m
_{2}
m
_{1}
is moving upward
(b)
m
_{1}
sinθ
=
2 m
_{2}
There will be no motion
(c)
If m
_{2}
exceeds a maximum value then mass m
_{2}
will start to move down and the friction force direction will be as shown in the free body diagram at right.
The friction force is equal to:
f = N
_{1}
μ = m
_{1}
g cosθ μ
From the forces diagram on mass m
_{1}
and m
_{2}
and assuming that the acceleration is to the downward direction of mass m
_{2}
we have.
On mass m
_{1}
ΣF
_{x}
= T
_{1}
− f − m
_{1}
g sinθ = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= m
_{2}
g − T
_{2}
= m
_{2}
a
_{2}
(2)
From the accelerations:
a
_{2}
= 2a
_{1}
(3)
From the pulley tension:
T
_{1}
= 2T
_{2}
(4)
Solving for a
_{1}
and T
_{2}
we get:
If we solve the case that m
_{1}
accelerates downward then the forces equations will be:
On mass m
_{1}
ΣF
_{x}
= m
_{1}
g sinθ − T
_{1}
− f = m
_{1}
a
_{1}
(5)
On mass m
_{2}
:
ΣF
_{y}
= T
_{2}
− m
_{2}
g = m
_{2}
a
_{2}
(6)
Eq. (3) and (4) are the same in both cases, solving the equations we get:
The range of m
_{2}
that the system will stay at rest is (in this case the accelerations a
_{1}
are equal to 0):