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Induced motion on rough incline (application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to induced motion on rough incline. The questions are categorized in terms of the characterizing features of the subject matter :

  • Nature of friction
  • Incline as mechanical device
  • Equilibrium

Nature of friction

Problem 1 : A force of 30 N is applied on a block of 10 kg, which is placed on an incline of angle, θ = sin - 1 3 / 5 θ = sin - 1 3 / 5 . If the coefficient of friction between block and incline is 0.75, then find the friction.

Figure 1: External force of 30 N acts up parallel to incline.
Block on a rough incline
Block on a rough incline (imq1.gif)

Solution : In order to find friction, we first need to know its direction. Here, the component of gravitational force along the incline is :

m g sin θ = 10 X 10 X sin sin - 1 3 5 = 10 X 10 X 3 5 = 60 N m g sin θ = 10 X 10 X sin sin - 1 3 5 = 10 X 10 X 3 5 = 60 N

The limiting friction is :

F S = μ m g cos θ = 0.75 X 10 X 10 X cos θ F S = μ m g cos θ = 0.75 X 10 X 10 X cos θ

In order to find the cosine, we use the triangle as shown. Here,

Figure 2: Cosine is ratio of base and hypotenuse.
Trigonometric ratio
Trigonometric ratio (imq2.gif)

⇒ cos θ = 5 2 − 3 2 5 = 4 5 ⇒ cos θ = 5 2 − 3 2 5 = 4 5

Hence,

⇒ F S = 0.75 X 10 X 10 X 4 / 5 = 60 N ⇒ F S = 0.75 X 10 X 10 X 4 / 5 = 60 N

Now, net component of external forces (without friction) along the incline is :

∑ F x = 60 − 30 = 30 N ∑ F x = 60 − 30 = 30 N

It means that block has the tendency to move down. Therefore, friction acts up. However, friction is not equal to limiting friction. The net component of external forces is 30 N and is less than limiting friction. Therefore, the static friction simply matches the net external force parallel to incline, which is 30 N.

f S = 30 N f S = 30 N

Incline as mechanical device

Problem 2 : A block can either be lifted vertically or pulled up along a rough incline of 30°. What should be coefficient of friction between block and incline so that pulling along the incline requires less force parallel to the incline.

Solution : In order to pull the block of mass “m” vertically, the force, “ F 1 F 1 ”, required is equal to the weight of the block,

F 1 = m g F 1 = m g

On the other hand, the net external force parallel to the incline should be greater than the limiting friction to move the block up the incline. The required force is :

Figure 3: Force required to pull up the block along an incline.
Block on a rough incline
Block on a rough incline (imq3.gif)

F 2 = m g sin θ + μ m g cos θ F 2 = m g sin θ + μ m g cos θ

For F 2 < F 1 F 2 < F 1 ,

⇒ m g sin θ + μ m g cos θ < m g ⇒ m g sin θ + μ m g cos θ < m g

⇒ sin θ + μ cos θ < 1 ⇒ sin θ + μ cos θ < 1

⇒ μ cos θ < 1 − sin θ ⇒ μ cos θ < 1 − sin θ

⇒ μ < 1 − sin θ cos θ ⇒ μ < 1 − sin θ cos θ

Putting value of the angle and evaluating, we have :

⇒ μ < 1 − sin 30 0 cos 30 0 ⇒ μ < 1 − sin 30 0 cos 30 0

⇒ μ < 1 − 1 / 2 3 2 ⇒ μ < 1 − 1 / 2 3 2

⇒ μ < 1 2 3 2 ⇒ μ < 1 2 3 2

⇒ μ < 1 3 ⇒ μ < 1 3

Note:

This example clearly demonstrate that (when friction is always present) the pulling along incline may not always involve lesser force as otherwise may be believed.

Equilibrium

Problem 3 : A block of mass 10 kg is placed on a rough incline plane of inclination 30° with an external force “F” applied as shown in the figure. The coefficient of friction between block and incline is 0.5. Find the minimum force “F” so that block does not slide on the incline.

Figure 4: Force required to pull up the block along an incline.
Block in equilibrium
Block in equilibrium (imq4.gif)

Solution : We first need to determine the direction of friction so that we could determine the minimum force required for block to remain stationary. Here, the gravitational force down the incline is :

m g sin θ = 10 X 10 X sin 30 0 = 10 X 10 X 1 2 = 50 N m g sin θ = 10 X 10 X sin 30 0 = 10 X 10 X 1 2 = 50 N

The external force “F” has no component along the incline. As such, the component of weight along the incline is the net force parallel to incline. Friction, therefore, acts upward. If external force “F” is not applied, the limiting friction is given by :

F S = μ m g cos θ = 0.5 X 10 X 10 X 3 2 = 25 3 N F S = μ m g cos θ = 0.5 X 10 X 10 X 3 2 = 25 3 N

As the downward gravitational force is greater than limiting friction, the block will slide down.

In order to restrict motion of block, the external force “F” should increase normal force, which in turn increases friction. Let the minimum force be “F” such that block does not side. In this limiting situation, the components of forces, parallel and perpendicular to incline, form balanced force systems. Free body diagram with external forces is shown in the figure. Considering "x" and "y" axes in parallel and perpendicular directions, we have :

Figure 5: Forces on the block.
Block in equilibrium
Block in equilibrium (imq5.gif)

∑ F x = m g sin θ = μ N ∑ F x = m g sin θ = μ N

∑ F y = m g cos θ + F = N ∑ F y = m g cos θ + F = N

Substituting for “N” in the first equation, we have :

⇒ m g sin θ = μ N = μ m g cos θ + μ F ⇒ m g sin θ = μ N = μ m g cos θ + μ F

⇒ F = m g sin θ − μ m g cos θ μ ⇒ F = m g sin θ − μ m g cos θ μ

⇒ F = 50 - 25 3 0.5 = 13.4 N ⇒ F = 50 - 25 3 0.5 = 13.4 N

Problem 4 : A block of mass “m” is placed on a rough incline of angle “θ”. The downward motion of the block is prevented by applying an upward force, “F”, parallel to incline. If a force of “2F” is required to initiate motion of the block in upward direction, then determine the coefficient of friction between block and the incline.

Solution : It is given that block has downward motion without any additional force. It means that net component of forces parallel to incline is in downward direction. The downward motion can be prevented if the additional force, parallel to the incline, is equal to the net downward component.

The additional force is equal to net downward component :

Figure 6: Forces on the block.
Block in equilibrium
Block in equilibrium (imq6.gif)

F = m g sin θ − μ m g cos θ F = m g sin θ − μ m g cos θ

In order to initiate motion in the upward direction, the additional external force should be equal to the net downward component of forces. As the block tends to move up, the friction, now, acts in the downward direction. Hence,

Figure 7: Forces on the block.
Block on an incline
Block on an incline (imq7.gif)

2 F = m g sin θ + μ m g cos θ 2 F = m g sin θ + μ m g cos θ

Subtracting first equation from the second, we have :

⇒ F = 2 μ m g cos θ ⇒ F = 2 μ m g cos θ

Putting this expression of “F” in the first equation and solving for “μ”,

⇒ 2 μ m g cos θ = m g sin θ − μ m g cos θ ⇒ 2 μ m g cos θ = m g sin θ − μ m g cos θ

⇒ 3 μ cos θ = sin θ ⇒ 3 μ cos θ = sin θ

⇒ μ = tan θ 3 ⇒ μ = tan θ 3

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