Problem 2 : A block of mass "m" is kept on a smooth incline of angle "θ" and mass "M". The incline is placed on a smooth horizontal surface. Find the horizontal force "F" required such that block is stationary with respect to incline.

Figure 2: The block is stationary on an accelerated incline.

Motion on accelerating incline

Solution : Since block is stationary with respect to incline, both elements of the system (incline and block) have same acceleration. As such, we can consider "block - incline" to constitute a combined system and treat them as one. As all surfaces are smooth, friction force is zero. Hence, acceleration of the system is given by :

a = F m + M a = F m + M

We can now analyze forces on the block to find force "F" for the given condition of block being stationary with respect to incline. The block is stationary with reference to incline, but it is accelerated with respect to ground. The condition of "being stationary" is, therefore, given in the context of accelerated frame of incline. Hence, it is more intuitive to carry out analysis in the non-inertial frame of reference.

Three forces on the block, including pseudo force, are shown here in the figure. Further, since two of the three forces i.e weight and pseudo forces are in horizontal and vertical directions, we choose our coordinates accordingly along these directions.

Figure 3: The block is stationary on an accelerated incline.

Forces on the block with pseudo force

The free body diagram of the block with components of forces along the axes is shown here.

Figure 4: The block is stationary on an accelerated incline.

Free body diagram in non-inertial frame

∑ F x = m a − N sin θ = 0 ∑ F x = m a − N sin θ = 0

⇒ N sin θ = m a ⇒ N sin θ = m a

∑ F y = N cos θ − m g = 0 ∑ F y = N cos θ − m g = 0

⇒ N cos θ = m g ⇒ N cos θ = m g

Taking ratio, we have :

⇒ a g = tan θ ⇒ a g = tan θ

Substituting for "a", we have :

F m + M g = tan θ F m + M g = tan θ

⇒ F = m + M g tan θ ⇒ F = m + M g tan θ

Problem 3 : A block of mass is placed on a rough incline of angle “θ”. The angle of incline is greater than the angle of repose. If coefficient of friction between block and incline is “μ”, then what should be acceleration of the incline in horizontal direction such that no friction operates between block and incline.

Figure 5: The angle of incline is greater than the angle of repose.

Block and incline system

Solution : In order to find the acceleration of the incline for “no friction”, we need to know the net component of forces parallel to the incline. The pulling force here is the component of gravitational force parallel to incline in the downward direction. It is given that the angle of incline is greater than angle of repose. It means that block has started moving downward and force of friction is equal to kinetic friction (about equal to limiting friction).

Let the mass of the block be “m”. The net component of force parallel to incline is acting downward and is given by :

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

The incline moves with a certain acceleration due to forces acting on it. We need to apply a pseudo force to apply Newton's second law of motion. It is clear that pseudo force should have a component along the incline in upward direction so that net component of force (without friction) is zero. This will ensure that there is no friction between interface of block and the incline.

Now, for the pseudo force to have a component along the incline upward direction, the acceleration of the incline should be towards left as indicated in the figure below.

Figure 6: The angle of incline is greater than the angle of repose.

Block and incline system

The free body diagram of the block with pseudo force is shown here.

Figure 7: The forces on the blocck.

Free body diagram

The components of gravitational force and pseudo force form balanced force system so that there is no net component of forces along the incline and hence there is no friction between the interface.

m g sin θ = m a cos θ m g sin θ = m a cos θ

⇒ tan θ = a g ⇒ tan θ = a g

a = g tan θ a = g tan θ towards left

Problem 4 : A block of mass "m" is placed on an incline of angle "θ" and mass "M", which is placed on a horizontal surface. All surfaces are friction - less. Find the acceleration of the block with respect to incline.

Figure 8: All surfaces are friction-less.

Block and incline system

Solution : Here, block and incline both have different accelerations. The block moves down along the incline. The incline, on the other hand, moves right as shown in the figure above. As such, we can not treat two elements as one body and find the their common accelerations with respect to ground.

Let " a B a B " , " a I a I " and " a BI a BI " respectively denote acceleration of block w.r.t ground, acceleration of the incline w.r.t ground and acceleration of the block w.r.t incline.

We first carry out force analysis on the "block" in the non-inertial frame of incline. Here, the forces on the block are (i) weight of the block, "mg", (ii) normal force due to underlying incline, " N 1 N 1 ", and (iii) Pseudo force, " m a I m a I " applied in the direction opposite to the acceleration of the incline. The free body diagram of the incline in non-inertial reference of incline is shown here.

Figure 9: Three forces, including pseudo force, act on the block.

Free body diagram of block in non-inertial frame

Force analysis on the block yields two relation,

∑ F x = m a I cos θ + m g sin θ = m a B I ∑ F x = m a I cos θ + m g sin θ = m a B I

⇒ a B I = a I cos θ + g sin θ ⇒ a B I = a I cos θ + g sin θ

∑ F y = N 1 + m a I sin θ = m g cos θ ∑ F y = N 1 + m a I sin θ = m g cos θ

It is clear from the first equation that we need to know " a I a I " i.e. the acceleration of incline with respect to ground to determine the value of " a BI a BI ". It is imperative that we carry out force analysis on the "incline" in the inertial frame of reference of ground.

The forces on the incline are (i) weight of the incline, "Mg", (ii) normal force due to overlying block, " N 1 N 1 ", and (iii) Normal force due to underneath horizontal surface, " N 2 N 2 ". The free body diagram of the incline in inertial ground reference is shown here.

Figure 10: Three forces act on the block.

Free body diagram of inline inertial frame

Force analysis of the incline yields two relations,

∑ F x = N 1 sin θ = m a I ∑ F x = N 1 sin θ = m a I

∑ F y = N 2 = N 1 cos θ + M g ∑ F y = N 2 = N 1 cos θ + M g

First of the equations is useful that can be combined with the analysis in non-inertial frame of reference. Substituting for " N 1 N 1 " in the earlier equation, we have :

⇒ m a I sin θ + m a I sin θ = m g cos θ ⇒ m a I sin θ + m a I sin θ = m g cos θ

⇒ a I = m g sin θ cos θ M + m sin 2 θ ⇒ a I = m g sin θ cos θ M + m sin 2 θ

We can now determine the required acceleration of the block with respect to incline by substituting the value of " a I a I " in the expression of " a BI a BI " :

⇒ a B I = m g sin θ cos 2 θ M + m sin 2 θ + g sin θ ⇒ a B I = m g sin θ cos 2 θ M + m sin 2 θ + g sin θ

⇒ a B I = m g sin θ cos 2 θ + M g sin θ + m sin 3 θ M + m sin 2 θ ⇒ a B I = m g sin θ cos 2 θ + M g sin θ + m sin 3 θ M + m sin 2 θ

⇒ a B I = g sin θ { M + m cos 2 θ + sin 2 θ } M + m sin 2 θ ⇒ a B I = g sin θ { M + m cos 2 θ + sin 2 θ } M + m sin 2 θ

⇒ a B I = M + m g sin θ M + m sin 2 θ ⇒ a B I = M + m g sin θ M + m sin 2 θ

Problem 5 : A block is placed on an incline, which is accelerated towards left at an acceleration equal to the acceleration due to gravity, as shown in the figure. The angle of incline is 30°. If friction is absent at all contact surfaces, then find the time taken by the block to cover a length of 4 m on the incline.

Figure 11: All surfaces are friction-less.

Block and incline system

Solution : Here, initial velocity of block is zero. We need to know the acceleration of the block with respect to incline in order to find the required time to cover a displacement of 4 m along the incline.

Figure 12: Forces on the block.

Block and incline system

Let “a” be the acceleration of the incline with respect to ground and “ a r a r ” be the acceleration of the block relative to the incline. We however do not know whether net component of forces parallel to incline is in up or down direction. The free body diagram of the block is superimposed on the block in the figure above. The forces on the block includes pseudo force (mg), acting towards right.

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

⇒ F x = m g cos 30 0 − m g sin 30 0 ⇒ F x = m g cos 30 0 − m g sin 30 0

⇒ F x = m g X 3 2 − m g X 1 2 ⇒ F x = m g X 3 2 − m g X 1 2

⇒ F x = 3 − 1 m g 2 ⇒ F x = 3 − 1 m g 2

Clearly, the net component of forces along the incline is positive. It means that block is moving up. The magnitude of acceleration with respect to incline in upward direction, is :

⇒ a r = 3 − 1 m g 2 m ⇒ a r = 3 − 1 m g 2 m

⇒ a r = 3 − 1 g 2 = 3.66 m / s 2 ⇒ a r = 3 − 1 g 2 = 3.66 m / s 2

Now, using equation of motion for constant acceleration, we have (initial velocity is zero) :

x = 1 2 a t 2 x = 1 2 a t 2

⇒ 4 = 1 2 X 3.66 X t 2 ⇒ 4 = 1 2 X 3.66 X t 2

⇒ t = 8 3.66 = 1.48 s ⇒ t = 8 3.66 = 1.48 s