Question

3. A block of mass m is placed on a rough horizontal surface . If u is the coefficient of static friction between. Then the maximum value of contact force between block and surface is(1) mg(2) mg√1+u²(3) 2mg√1+u²(4) mg √1+u²/2​

Answer 1

Answer:

Weight of the body = mg.

Let the Horizontal Force be L.

∴ Resultant Force = √(L² + m²g²)

Now, we know that,

 L ≤ friction (because body is not moving)

∴ L = friction = μmg (Maximum)

⇒ F ≤ √(μ²m²g² + m²g²)

∴ F ≤ mg√(μ²+ 1)

Now, This F will make angle θ with the Horizontal which will give FCosθ.

∴ F = mg - F Sinθ

∵ θ should always be less than 90° therefore, it can always be less than 1.

∴ F ≥ mg

∴ Mg ≤ F ≤ Mg√(1+µ2)

Hence, Option (c). is correct.

Answer 2

Therefore the maximum value of contact force between block and surface is mg√ ( 1 + u² ). ( Option-2 )

Given:

Mass of the block = m kg

Coefficient of static friction between the block and the surface = u

Acceleration due to gravity = g m/s²

To Find:

The maximum value of contact force between the block and the surface.

Solution:

The given question can be solved as shown below.

Given that,

Mass of the block = m kg

Coefficient of static friction between the block and the surface = u

Acceleration due to gravity = g m/s²

Maximum value of contact force is obtained when the body just starts to move.

Contact force = Resultant of Normal force and Frictional force

⇒ Contact Force = R = √ ( F² + N² )

Where F = Frictional Force = uN

N = Normal Force

On substituting the values in the formula,

⇒ Contact Force = R =√ ( uN )² + N²

⇒ R = N√ ( 1 + u² )

If the body is resting on an horizontal plane, then the Normal force = mg

So Contact Force = R = mg√ ( 1 + u² )

Therefore the maximum value of contact force between block and surface is mg√ ( 1 + u² ).

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