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.

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

Here N_{A }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., N_{A} = N_{B}

**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., N_{A} > N_{B}

**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,

N_{B} = 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.

- 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 :**

- mg (acting through the center of mass of the body)
- Horizontal force (F) applied
- Friction force (opposite to the applied force F)

**Under equilibrium of block :**

Using FBD (II)

F_{x }= 0

f = F

For normal force

F_{y} = 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.

- 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.

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

- 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 \(\mu_{mg}\).

- 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 :**

- 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 **

- 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)

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

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

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

- 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.

- 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., \(\mu>\dfrac{b}{2h}\)

A Topple before slipping

B Slip before toppling

C Insufficient information

D Both conditions occur together

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