Informative line

Toppling

Learn normal contact force read by weighing machine in free body diagram. Practice to find the minimum value of horizontal force required for the slipping of block and FBD of block at equilibrium pushed horizontally.

Man on Two Weighing Machine

  • Consider a man of mass m, standing with his feet on two weighing machines.

Free Body Diagram

Here NA and NB are normal contact force read by weighing machine A and B respectively.

Case 1: No horizontal force on man

  • Under no horizontal force, the readings of both weighing machines are same.

i.e., NA = NB

Case 2: Horizontal force applied

  • Consider a horizontal force acting on a man, as shown in figure.
  • The reading of weighing machine A increases and that of B decreases.

i.e., NA > NB

Case 3: Horizontal force increased

  • As the value of force is increased, there will be a situation where the contact of his feet with weighing machine B breaks.

Under this condition,

NB = 0

  • In this condition, the normal contact force at the left foot, i.e., at weighing machine A increases to its maximum value.

Conclusion 

The normal force shifts in the direction of the push, i.e., in the direction of the force applied on it.

Illustration Questions

A block of mass m, kept on a horizontal rough surface, is pushed by a horizontal force, as shown in figure. Which one of the following diagrams represents the position of normal force correctly?

A

B

C

D

×

When no horizontal force acts,

FBD of block

image

On application of horizontal force, the normal shifts to adjust itself in the direction of push.

image

Normal force is the resultant of interaction force of all the particles of the two surfaces in contact. As horizontal force is applied, the surfaces push against each other. The particles on the left of C push against each other more strongly than those to the right of C. This results in shifting of net resultant normal towards left.

A block of mass m, kept on a horizontal rough surface, is pushed by a horizontal force, as shown in figure. Which one of the following diagrams represents the position of normal force correctly?

image
A image
B image
C image
D image

Option B is Correct

FBD of Block at Equilibrium Pushed Horizontally

  • Consider a block of mass m, kept on a horizontal rough surface with friction coefficient \(\mu\), is pushed by a force F as shown.

  • Forces acting on the block are :
  1. mg (acting through the center of mass of the body)
  2. Horizontal force (F) applied
  3. Friction force (opposite to the applied force F)

  • Under equilibrium of block :

 Using FBD (II)

F= 0

f = F

For normal force

Fy = 0

N – mg = 0

N = mg   [ using FBD II ]

FBD for center of mass ( II )

  • The point of application of normal force can be determined by applying the condition of rotational equilibrium \((\tau =0)\).
  • Under equilibrium condition, torque about all the points is zero.
  • The point through which maximum number  of unknown forces pass, is a good point to calculate torque in condition of equilibrium.

Illustration Questions

A block of mass m, kept on a horizontal rough surface with friction coefficient \(\mu\), is pushed by a force F, as shown. The FBD for center of mass is -

A

B

C

D

×

Forces acting on the block are 

mg (acting through the center of mass of the body)

Horizontal force (F) applied

Friction force (opposite to the applied force F)

For center of mass, the point of application of force is not required. So, all the forces can be drawn at the same point. 

 

image

A block of mass m, kept on a horizontal rough surface with friction coefficient \(\mu\), is pushed by a force F, as shown. The FBD for center of mass is -

image
A image
B image
C image
D image

Option B is Correct

Minimum Force Required to Slip the Block

  • Consider a block of mass m, kept on horizontal rough surface with friction coefficient \(\mu\), is pushed by a force F, as shown.

Forces acting on the block are

mg (acting through the center of mass of the body)

Horizontal force (F) applied

Friction force (opposite to the applied force F)

When a horizontal force is applied, N shifts in the direction of force. 

For center of mass, the point of application of force is not required. So, all the forces can be drawn at the same point.

For Slipping 

  • FBD (II) is more appropriate as slipping involves translational motion, i.e., motion of center of mass.

Condition for Slipping 

  • For slipping, the applied horizontal force should be greater than the limiting value of friction.

\(F–f_{max}>o\)

or, \(F>f_{max}\)

or, \(F>\mu N\)

or, \(F>\mu\, mg\)            \(\therefore \,\,\,[As\,\Sigma Fy=0,\,N–mg=0,\,N=mg]\)

So, minimum force required to slip is \(\mu_{mg}\).

Illustration Questions

A block of mass \(m=4\,kg\) is kept on a horizontal rough surface. If coefficient of friction is  \(\mu=0.2\). Find the minimum value of horizontal force required for the slipping of block. \([\text{given}\,\,\,g=10\,m/sec^2]\) 

A 7 N

B 6 N

C 8 N

D 5 N

×

To determine the minimum force required to slip the block, the acceleration of center of mass is to be analyzed.

For slipping, FBD II is important as slipping involves translational motion of the body.

image

Condition for slipping

\(F–f_s>0\)

or, \(F>f_s\)

If F is greater than the limiting value of static friction, the block will slip.

\(\therefore\,F_{min}=(f_s)_{lim}\)

or, \(F_{min}=\mu N\)

Also, \(F_y=0 \)    [from II]

\(N–mg=0\)

\(N=mg\)

So, \(F_{min}=\mu mg\)

Given, \(m=4\,kg,\,\mu=0.2,\,g=10\,m/sec^2\)

\(F_{min}=\mu{mg}\)

or, \(F_{min}=0.2×4×10\)

\(F_{min}=8\,N\)

A block of mass \(m=4\,kg\) is kept on a horizontal rough surface. If coefficient of friction is  \(\mu=0.2\). Find the minimum value of horizontal force required for the slipping of block. \([\text{given}\,\,\,g=10\,m/sec^2]\) 

A

7 N

.

B

6 N

C

8 N

D

5 N

Option C is Correct

Minimum Force Required for Toppling

  • Consider a block of mass m, kept on a horizontal rough surface with friction co-efficient  \(\mu\), is pushed by a force F, as shown.

Forces acting on the block are :

  1. mg (acting through the center of mass of the body)
  2. Horizontal force (F) applied
  3. Friction force (opposite to applied force F)
  4. Normal force - When the block is just about to topple, the normal force shifts to the left extreme.

FBD of block 

  • For toppling of block, FBD I is required.
  • For toppling, rotation of block needs to be analyzed.
  • As the toppling occurs about point P,

\(\tau_p>0\)

\(\tau_F+\tau_f+\tau_N+\tau_{mg}>0\)

\(F\times\,h–mg×\dfrac{b}{2}>0\)   ...(1)

[f and N passes through P, So, \(\tau _f\) and \(\tau _N=0\) because perpendicular distance = 0]

From (1)

\(\therefore\,\,\,Fh>mg\dfrac{b}{2}\)

\(F>\dfrac{mgb}{2h}\)

\(F_{min}=\dfrac{mgb}{2h}\)

Illustration Questions

A block of mass \(m=6\,kg\), kept on a horizontal rough surface with friction coefficient  \(\mu\), is pushed by a force F, as shown. Calculate the minimum force required to topple the block. Given (b = 6 m, h = 1 m)

A 100 N

B 180 N

C 50 N

D 150 N

×

Forces acting on the block are 

mg (acting through the center of mass of the body)

Horizontal force (F) applied

Friction force (opposite to applied force F)

Normal force - When the block is just about to topple, the normal force shifts to the left extreme.

FBD of block 

image

For toppling, rotation of block needs to be analyzed.

As the toppling occurs about point P,

\(\tau_p>0\)

\(\tau_F+\tau_f+\tau_N+\tau_{mg}>0\)

\(F\times\,h–mg×\dfrac{b}{2}>0\)   ...(1)

[f and N passes through P, So, \(\tau_ f\) and \(\tau _N=0\) because perpendicular distance = 0]

From (1)

\(\therefore\,\,\,Fh>mg\dfrac{b}{2}\)

\(F>\dfrac{mgb}{2h}\)

\(F_{min}=\dfrac{mgb}{2h}\)

image

\(F_{min}h=mg\dfrac{b}{2}\)

or, \(F_{min}\times1=6×10×\dfrac{6}{2}\)

or, \(F_{min}=180\,N\)

A block of mass \(m=6\,kg\), kept on a horizontal rough surface with friction coefficient  \(\mu\), is pushed by a force F, as shown. Calculate the minimum force required to topple the block. Given (b = 6 m, h = 1 m)

image
A

100 N

.

B

180 N

C

50 N

D

150 N

Option B is Correct

What will Happen First : Toppling or Slipping?

  • Consider a block of mass m, kept on a horizontal rough surface with friction coefficient \(\mu\), is pushed by a force F, as shown.
  • The force is increased gradually, till the block topples or slips.

Condition of Slipping and Toppling 

  • Minimum force required for slipping is \(\mu\,mg\)
  • Minimum force required for toppling is \(\dfrac{mgb}{2h}\)

Case 1- When minimum force required for toppling is greater than that required for slipping, the block will slip first.

i.e., \(F_{min}\) for toppling > \(F_{min}\) for slipping

or, \(\dfrac{mgb}{2h}>\mu mg\)

or, \(\mu<\dfrac{b}{2h}\)

Case 2- If minimum force required for toppling is less than that required for slipping, the block will topple.

i.e., \(F_{min}\) for toppling < \(F_{min}\) for slipping

or, \(\dfrac{mgb}{2h}<\mu mg\)

i.e., \(\mu>\dfrac{b}{2h}\)

Illustration Questions

A block of height H and width \(b=2\,H\) is kept on a rough horizontal surface. The coefficient of friction between the surfaces is \(\mu=1\). A horizontal force is applied to the block at height \(h=\dfrac{3\,H}{5}\). As the force is increased gradually, the block will:

A Topple before slipping

B Slip before toppling

C Insufficient information

D Both conditions occur together

×

Condition of slipping and toppling 

Minimum force required for slipping is \(\mu\,mg\)

Minimum force required for toppling is \(\dfrac{mgb}{2h}\)

 

image

Case 1- When minimum force required for toppling is greater than that required for slipping, the block will slip first.

i.e., \(F_{min}\) for toppling > \(F_{min}\) for slipping

or, \(\dfrac{mgb}{2h}>\mu mg\)

or, \(\mu<\dfrac{b}{2h}\)

 

image

Case 2- If minimum force required for toppling is less than that required for slipping, the block will topple.

i.e., \(F_{min}\) for toppling < \(F_{min}\) for slipping

or, \(\dfrac{mgb}{2h}<\mu mg\)

i.e., \(\mu>\dfrac{b}{2h}\)

 

image

Given  \(h=\dfrac{3\,H}{5}\,,\, b=2H\)

So, \(\dfrac{b}{2h}\)

\(=\dfrac{2\,H×5}{2×3\,H}\)

\(=\dfrac{5}{3} \)

\(=1.66\)

 

image

Since, \(\mu<\dfrac{b}{2h}\Rightarrow\,\mu<1.66\)

So, block will slip first

image

A block of height H and width \(b=2\,H\) is kept on a rough horizontal surface. The coefficient of friction between the surfaces is \(\mu=1\). A horizontal force is applied to the block at height \(h=\dfrac{3\,H}{5}\). As the force is increased gradually, the block will:

A

Topple before slipping

.

B

Slip before toppling

C

Insufficient information

D

Both conditions occur together

Option B is Correct

Illustration Questions

A block of mass m, height h and width b is kept on a sufficiently rough inclined surface so that no slipping occurs. If the angle of inclination is increased gradually, at what angle will the block topple?

A \(tan^{-1}\left(\dfrac{b}{h}\right)\)

B \(tan^{-1}\left(\dfrac{h}{b}\right)\)

C \(sin^{-1}\left(\dfrac{b}{h}\right)\)

D \(cos^{-1}\left(\dfrac{h}{b}\right)\)

×

At equilibrium \(\Sigma y=0\)

\(N=mg\,cos\theta\)

as no slipping occurs

\(F_x=0\)

\(f=mg\,sin\theta\)

image

Torque at C 

\(\tau_c=\tau_{mg}+\tau_f+\tau_N\)

\(\tau_c=0+mg\,sin\theta×\dfrac{h}{2}–mg\,cos\theta×\dfrac{b}{2}\)

[friction rotates block in anticlockwise direction (+ve) and normal in clockwise direction (–ve)]

image

For block to topple 

\(\tau_c\) is anticlockwise | positive |\(\odot\)

\(\tau_c>0\)

\(mg\,sin\theta×\dfrac{h}{2}=mg\,cos\theta×\dfrac{b}{2}\)

\(tan\theta=\dfrac{b}{h}\)

\(\Rightarrow\,\theta=tan^{-1}\left(\dfrac{b}{h}\right)\)

A block of mass m, height h and width b is kept on a sufficiently rough inclined surface so that no slipping occurs. If the angle of inclination is increased gradually, at what angle will the block topple?

image
A

\(tan^{-1}\left(\dfrac{b}{h}\right)\)

.

B

\(tan^{-1}\left(\dfrac{h}{b}\right)\)

C

\(sin^{-1}\left(\dfrac{b}{h}\right)\)

D

\(cos^{-1}\left(\dfrac{h}{b}\right)\)

Option A is Correct

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