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Motion in accelerated frame (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 motion in accelerated frame. The questions are categorized in terms of the characterizing features of the subject matter :

  • Motion in a lift
  • Motion of an incline on smooth surface
  • Motion on an incline

Motion in a lift

Problem 1 : What should be the acceleration “a” of the lift so that a block of mass “m” placed on the floor exerts a normal force of "mg/3" on it.

Figure 1: The lift is moving with acceleration.
A block on the floor of a lift
 A block on the floor of a lift  (fa2a.gif)

Solution : The normal force on the floor is less than that exerted by the block when the lift is stationary. It means that lift is accelerating in downward direction.

In order to find acceleration of the lift, we analyze force on the block in the non-inertial frame of lift. The forces of the block are (i) its weight, “mg”, acting downward (ii) Normal force, “N”, acting upward and (iii) pseudo force, “ma” acting upward in the direction opposite to the direction of lift's acceleration.

The free body diagram of the block in the frame of lift is shown in the figure. Analyzing force in the vertical direction, we have :

Figure 2: The forces on the block.
Free body diagram
 Free body diagram  (fa1.gif)

N + m a = m g N + m a = m g

Substituting for normal force as given in the question,

M g 3 + m a = m g M g 3 + m a = m g

a = g g 3 = 2 g 3 a = g g 3 = 2 g 3

Motion of an incline on smooth surface

The problems here are related to the motion of incline, which is accelerated. The incline serves as the accelerated frame for the motion of block. In this sense, these problems do not exactly pertain to the motion in accelerated frame. We shall, therefore, analyze the motion of incline in inertial ground reference. Problems, involving motion of block in the accelerated frame of incline, are included in the next module.

Problem 2 : In the arrangement shown in the figure, “N” is the normal force applied by the block on the incline of angle “θ”. If all surfaces are smooth, then find the acceleration of the incline.

Figure 3: All the surfaces are smooth.
Block and incline system
 Block and incline system  (fa3.gif)

Solution : Since all surfaces are smooth, the incline will accelerate due to the normal force applied by the block.

The forces on the incline are (i) its weight, “mg”, (ii) Normal force, “N”, applied by the overlying block and (iii) normal force, “ N 1 N 1 ” due to the surface underneath. There is no motion of incline in the vertical direction. Thus, we need to analyze components of forces in horizontal direction only to find the acceleration in that direction.

Figure 4: Forces on the incline.
Block and incline system
 Block and incline system  (fa4.gif)

F x N sin θ = M a F x N sin θ = M a

a = N sin θ M towards right a = N sin θ M towards right

Problem 3 : In the arrangement shown in the figure, “N” is the normal force applied by the block on the incline of angle “θ”. If all surfaces are smooth, then find normal force on the incline, as applied by the horizontal surface underneath.

Figure 5: All the surfaces are smooth.
Block and incline system
 Block and incline system  (fa3.gif)

Solution : Since all surfaces are smooth, the incline will accelerate due to the normal force applied by the block.

The forces on the incline are (i) its weight, “mg”, (ii) Normal force, “N”, applied by the overlying block and (iii) normal force, “ N 1 N 1 ” due to the surface underneath. There is no motion of incline in the vertical direction. Hence, components of forces in vertical direction constitute a balanced force system.

Figure 6: Forces on the incline.
Block and incline system
 Block and incline system  (fa4.gif)

F y = N 1 M g + N cos θ = 0 F y = N 1 M g + N cos θ = 0

N 1 = M g + N cos θ N 1 = M g + N cos θ

Motion on an incline

Problem 4 : A pendulum bob of mass 2 kg hangs from the ceiling of a train compartment. The train is accelerating up an incline terrain at 5 m / s 2 m / s 2 , making an angle 30° with horizontal. Find the angle that the pendulum bob makes with the normal to the ceiling.

Figure 7: A pendulum bob hangs from the ceiling of a train compartment.
Pendulum
 Pendulum  (fa5.gif)

Solution : Let the pendulum bob be at an angle “α” with the normal to the ceiling of the compartment as shown in the figure below. The bob is in a stationary position in the non-inertial frame of the accelerating compartment. We can use this fact in the non-inertial frame. As such, we shall carry out force analysis in the accelerated frame of the compartment.

The forces on the bob are (i) its weight “mg”, acting in vertically downward direction (ii) Tension in the string, “T”, making an angle “α” with the normal to the ceiling and (iii) pseudo force, “ma”, acting down the incline in the direction opposite to the direction of acceleration of frame of reference.

Figure 8: Forces on pendulum bob.
Pendulum
 Pendulum  (fa6.gif)

As pseudo force is along the incline, we select coordinate axes parallel and perpendicular to the incline terrain as shown in the figure. FBD as superimposed on the diagram is shown in the figure.

F x T sin α = m a + m g sin 30 0 F x T sin α = m a + m g sin 30 0

T sin α = 2 X 5 + 2 X 10 2 = 20 T sin α = 2 X 5 + 2 X 10 2 = 20

F y T cos α = m g cos 30 0 F y T cos α = m g cos 30 0

T cos α = 2 X 10 X 3 2 = 10 3 T cos α = 2 X 10 X 3 2 = 10 3

Taking ratio of two equations, we have :

tan α = 2 3 tan α = 2 3

From the analysis "x" direction, the tension in the string is :

T = 20 sin α T = 20 sin α

We need to know "sinα" in order to evaluate the expression of tension. Using the ratio of tangent, we can determine the sine of the angle :

Figure 9: Sine of the angle.
Trigonometric ratio
 Trigonometric ratio  (fa7.gif)

sin α = 2 { 2 2 + 3 2 } = 2 7 sin α = 2 { 2 2 + 3 2 } = 2 7

Putting in the expression of tension, ”T”,

T = 20 7 2 = 10 7 N T = 20 7 2 = 10 7 N

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