Friction solved problems

Friction and pulleys solved problems

Print notes about calculating pulley and friction problems

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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 / s² ]
A force of 1 Newton is equal to	N = 1 [kgm m / s² ]
Then if the weight is given it equals	W [kgf ] = 9.8 N

There are two friction coefficients the static μ_s and 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.

Direction of acceleration and the friction force

In many questions we have to decide the direction of the movement, it is important because the direction of the friction force is always opposite to the movement direction.

If we look at the figure at left the direction of the movement is to the right as the direction of a₁ the equilibrium equation will be:

ΣF_x = F − m g sin θ − f = m a

and the acceleration is:

Acceleration value movement to the right

Opposite direction of acceleration and friction force

If the movement 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 from 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.

Acceleration of two masses connected by pulley

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 travelled is given by: x = v₀ t + a t² / 2

At rest v₀ = 0 and the acceleration is: a = 2 x / t²

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: a₁ = 2 a₂

For system rest we need that the acceleration will be 0. From the equation v_t = v₀ + a t we get the result v_t = v₀

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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:

Friction coefficient of a tilted surface

We can see that the friction coefficient for rest condition must be equal or greater than the tangent of the slope μ > tan θ.

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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 = v₀ + 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/s²

And the time until mass m will stop is according to the equation: v_t = v₀ + a t where v_t = 0 and v₀ = 38.5 m/s

t = − v₀ / a = − 38.5 / − 1.96 = 19.64 sec

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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 acceleration of mass M if the friction coefficients between the two masses is μ₁ and μ₂ between mass M and the lower surface.

Friction force f₁ is:	f₁ = mgμ₁
Friction force f₂ is:	f₂ = (M + m)gμ₂

Mass M is affected by both frictions f₁ and f₂ and is:

f_M = f₁ + f₂ = mgμ₁ + (M + m)gμ₂

From the forces diagram we can derive the equilibrium equations in the x direction on both masses:

On mass m:	ΣF_x = T₁ − mgμ₁ = 0	(1)
On mass M:	ΣF_x = F − f₁ − f₂ = Ma	(2)

Maximum force F that can be applied to mass M without sliding is:

F <= mgμ₁ + (M + m)gμ₂

From eq (1) T₁ can be found:

T₁ = mgμ₁

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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) :

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Free body diagram acceleration to the right

Figure 1

Free body diagram acceleration to the left

Figure 2

M		kg
m		kg
θ		deg
μ
a		m/s₂
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 these 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 assuming this time that the acceleration is downward see the case of (Figure 2).

Notice that we got a different value for the acceleration, the conclusion is that mass M is sliding 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.

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Free body diagram when acceleration is to the right

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)

Acceleration when motion is to the right

System is at rest when:

m(sinβ − μ_2scosβ) − 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 < tanα

μ_2s < tanβ

Free body diagram when acceleration is to the left

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/s²
T		N

No friction move:

Each mass move:

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

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 criteria (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 criterion (c) and (d) to verify is motion occurs:

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:

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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 μ₁ μ₂ and the slope angle θ are known.

Notice that the two masses are moving in the opposite directions so the friction forces are:

Friction forces on mass M are:	f₁ = (M + m)gμ₁ cosθ
Friction forces on mass m are:	f₂ = mgμ₂ 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 − f₂ − mg sinθ = ma	(1)
On mass M	ΣF_x = Mg sinθ − f₁ − f₂ − T = Ma	(2)

Substitute the values of f₁ and f₂ into eq. 1 and 2 and eliminating T from both equations we get a:

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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:

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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 acceleration 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 = Ma₁	(1)
Forces on mass m:	ΣF_y = mg − T₁ = ma₂	(2)
From acceleration:	a₁ = 2a₂	(3)
From the pulley:	T₁ = 2T	(4)

We got four equations with four unknowns T T₁ a₁ and a₂ , solving the equations we get:

The accelerations are:
The tensions are:

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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 = Ma₁	(1)
Forces on mass m:	ΣF_y = mg − T = ma₂	(2)
Acceleration’s balance:	a₂ = 2a₁	(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 a₁ a₂ and T we get:

And the tension in the rope is:

The tension T is uniform along the cable.

The value of T₁ is:

T₁ = 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.

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m₁		kg
m₂		kg
m₃		deg
a₁		m/s²
a₂		m/s²
a₃		m/s²
T₁		N
T₂		N

Acceleration directions
m₁	m₂	m₃

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 a₁ a₂ and a₃ . Now we can write the forces acting on each mass as:

ΣF_y on mass m₁	m₁ g − T₁ = m₁ a₁	(1)
ΣF_y on mass m₂	m₂ g − T₂ = m₂ a₂	(2)
ΣF_y on mass m₃:	m₃ g − T₂ = m₃ a₃	(3)
ΣF_y on pulley B:	T₁ = 2 T₂	(4)

We can see that we have 5 unknowns and only 4 equations so we have to derive another equation from the accelerations, a₁ is equal to the acceleration of pulley B which should be half the accelerations of a₂ and a₃ but in negative sign because when a₁ is going down pulley B is going up, but we chose the accelerations in the downward direction.

Σ Accelerations:	a₁ = −(a₂ + a₃) / 2	(5)
	a₃ = − 2a₁ − a₂

The minus sign is because a₁ is in the opposite direction to a₂ and a₃. Substitute the value of T₁ from eq. (4) and a₃ from eq. (5) to eq. (1) (2) and (3) to get the matrix form:

Solving by Cramer's rule we have:

D = 2 (m₂m₃ + m₂m₃) + m₁(m₃ + m₂) = m₁m₂ + m₁m₃ + 4m₂m₃

After division of the numerator and the denominator by m₁m₂m₃ we got another form for T₂

If all masses are the same and equal to M then:


	T₂ = 2 g M / 3

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from the forces diagram we can write the equilibrium equations:

On mass m₁	ΣF_x = T₁ = f₁ = m₁ g μ₁	(1)
On mass m₂	ΣF_x = T₂ − f₁ − f₂ = m₂ a
	T₂ − m₁ g μ₁ − (m₁ + m₂) μ₂ g = m₂ a	(2)
On mass m₃	ΣF_y = m₃ g − T₂ = m₃ a	(3)
		(4)

From eq. (3) T₂ can be found

T₂ = m₃(g − a) =

Q:	Given three masses m₁ equal to m, mass m₂ equal to 3m and m₃ equal to 2m, μ₁ = 0 and masses m₂ and m₃ are moving at a constant velocity. Find the value of μ₂, T₁ and T₂
S:	For steady velocity we need that a = 0 From the acceleration equation (4) we have: m₃ − m₁ μ₂ − m₂ μ₂ = 0 Because there is no friction between surfaces of masses m₁ and m₂ then: T₁ = 0 Because the acceleration is 0 the tension T₂ in cable m₃ is: T₂ = m₃ g

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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 m₁ and mass m₂ as upward and the acceleration of m₂ downward.

From the forces diagram on the masses, we have.

On mass m₁	ΣF_y = T − m₁ g = m₁ a₁	(1)
On mass m₂:	ΣF_y = m₂ g − 2T = m₂ a₂	(2)
On mass m₃:	ΣF_y = T − m₃ g = m₃ a₃	(3)
From accelerations:		(4)

Note: when mass m₁ is moving x₁ distance upward and mass m₃ a distance of x₃ upward, the contribution to the displacement of mass m₂ is the sum of half of this displacements x₂ = (x₁ + x₃) / 2 in the downward direction.

The equations with the unknowns T, a₁, a₂ and a₃ can be solved by Cramer's rule or by direct substitution.

Write the equations in the matrix form

D = − m₂m₃ − 4m₁m₃ − m₁m₂ = − (m₁m₂ + m₂m₃ + 4m₁m₃)

Find the acceleration and the tension in the rope if all the masses are equal to m₁ = m₂ = m₃ = m

T = 2 m g / 3

a₁ = − g / 3

a₂ = − g / 3

a₃ = − g / 3

T = 2 m / 3

Notice that if m₁ = m₃ = m and m₂ = 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.

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Three masses are connected with a rope of 0 mass as shown in the figure, a force F is apply to mass m₃ , friction coefficient μ if present between mass m₁ 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 analyse the relationship of the equations:

m₃ > m₂ + m₁ sinθ	Motion is m₃ down	(a)
m₃ < m₂ + m₁ sinθ	Motion is m₃ up	(b)
m₃ = m₂ + m₁ sinθ	No motion	(c)

Assume that the acceleration of mass m₃ is downward then:

On mass m₁	ΣF_x = T₁ − μ m₁ g cos θ − m₁ g sinθ = m₁ a	(1)
On mass m₂	ΣF_y = T₂ − m₂ g − T₁ = m₂ a	(2)
On mass m₃	ΣF_y = m₃ g + F − T₂ = m₃ a	(3)

The same equations when the acceleration is to the left are:

On mass m₁	ΣF_x = m₁ g sinθ − T₁ − μ m₁ g cosθ = m₁ a	(4)
On mass m₂	ΣF_y = m₂ g + T₁ − T₂ = m₂ a	(5)
On mass m₃	ΣF_y = T₂ − m₃ g − F = m₃ a	(6)

We got three equations with three unknowns T₁ T₂ 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 needed to apply when a = 0, when the motion is intended to be to the left is:

F = g ( m₁ sinθ + m₂ − m₃ ) − m₁μ_s g cosθ

The maximum force F needed to apply when a = 0, when the motion is intended to be to the right is:

F = g ( m₁ sinθ + m₂ − m₃ ) + m₁μ_s g cosθ

In the case when F = 0 mass m₃ that keeps the system at rest is in the range of:

m₁ ( sinθ − μ cosθ ) + m₂ < m₃ < m₁ ( sinθ + μ cosθ ) + m₂

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)

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Three masses are connected to each other by ropes of 0 mass, a force F is applied to mass m₃. 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:

f₁ = m₁ g μ

f₂ = m₂ g μ

f₃ = m₃ g μ

ΣF_x = T₁ − m₁ g μ = m₁ a	(1)
ΣF_x = T₂ − T₁ − m₂ g μ = m₂ a	(2)
ΣF_x = F − T₂ − m₃ g μ = m₃ a	(3)

We got three equations with three unknowns T₁ T₂ and a

According to Cramer's rule the value of the coefficient’s determinant is D = − (m₁ + m₂ + m₃)

Another way to solve the problem is to look on the three masses as one mass equal to M = m₁ + m₂ + m₃

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 T₁ and T₂

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Two masses m₁ and m₂ are located on the floor the masses then connected 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 m₁ > m₂.

We mark the accelerations of the masses as a₁ and a₂, from the free body diagram we get:

On mass m₁:	ΣF_y = T − m₁ g = m₁ a₁	(1)
On mass m₂:	ΣF_y = T − m₂ g = m₂ a₂	(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 a₁ and a₂ can not be negative (opposite to the direction shown) because of the floor there are some unique cases:

Small mass m₂	F < 2 m₂ g	The force F is not enough to lift any one of the masses
Bigger mass m₁	2 m₂ g < F < 2 m₁ g	The force F will lift the small mass but not the bigger mass
	2 m₁ g < F	The force F will lift both masses

For example, if mass m₁ = 30 kg and m₂ = 20 kg then find the accelerations when F = 100N, F = 400N and F = 1000N.

F [N]	a₁ [m/s²]	a₂ [m/s²]	Notes
100N	100 / 30 − 9.8 = − 6.5	100 / 20 − 9.8 = − 4.8	No lift of any mass a₁ = 0 a₂ = 0
200N	200 / 30 − 9.8 = − 3.1	200 / 20 − 9.8 = 0.2	Only small mass is lifted a₁ = 0
1000N	400 / 30 − 9.8 = 3.5	400 / 20 − 9.8 = 10.2	Both masses move upward

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m₁		kg
m₂		kg
μ
F		N
θ		deg
a		m/s₂
T		N
f		N

Mass m₁ and m₂ 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 m₁ is: f = N μ = (m₁ g + F sin θ) μ.

In order to determine the possible direction of the motion we have to check the forces acting on mass m₁ without friction force.

if F cosθ > m₂ g	m₁ moves to the left	(a)
if F cosθ < m₂ g	m₁ 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 m₁ is to the left, from the free body diagram we have

assume

On m₁	F cosθ − f − T = m₁ a	(1)
On m₂	T − m₂ g = m₂ a	(2)

From eq. (1) and (2) we can find the values of a and T

If we assume that the acceleration of mass m₁ is to the right direction the equilibrium equations will be as follows:

On m₁	T − F cosθ − f = m₁ a	(3)
On m₂	m₂ g − T = m₂ 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 > m₂ g	m₁ moves to the left	(c)
If	F cosθ + f < m₂ g	m₁ 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 m₁ is μ = 0.4. Find the friction force acting on m₁ if the force equal 250N and θ = 30 degree and m₁ = 10kg and m₂ = 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 m₂ g and F * cosθ

	*f = m₂ g − F cosθ** = 30 * 9.8 − 250 * cos30 = 77.5 N
b) maximum force F	F = g(100.4 + 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 / s²

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Two masses m₁ and m₂ are connected through 2 pulleys as shown in the figure the friction of m₁ 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 analyse the forces on the pulley by neglecting the friction force, it is easy to see that if:

m₁ sinθ > 2 m₂	m₁ is moving downward	(a)
m₁ sinθ < 2 m₂	m₁ is moving upward	(b)
m₁ sinθ = 2 m₂	There will be no motion	(c)

If m₂ exceeds a maximum value then mass m₂ 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 = N₁ μ = m₁ g cosθ μ

From the forces diagram on mass m₁ and m₂ and assuming that the acceleration is to the downward direction of mass m₂ we have.

On mass m₁	ΣF_x = T₁ − f − m₁ g sinθ = m₁ a₁	(1)
On mass m₂:	ΣF_y = m₂ g − T₂ = m₂ a₂	(2)
From the accelerations:	a₂ = 2a₁	(3)
From the pulley tension:	T₁ = 2T₂	(4)

Solving for a₁ and T₂ we get:

If we solve the case that m₁ accelerates downward then the forces equations will be:

On mass m₁	ΣF_x = m₁ g sinθ − T₁ − f = m₁ a₁	(5)
On mass m₂:	ΣF_y = T₂ − m₂ g = m₂ a₂	(6)

Eq. (3) and (4) are the same in both cases, solving the equations we get:

The range of m₂ that the system will stay at rest is (in this case the accelerations a₁ are equal to 0):