Question

If a ladder weighing 250n is placed against a smooth

vertical wall having coefficient of friction between it

and floor is 0.3, then what is the maximum force of

friction available at the point of contact between the

ladder and the floor

Answer 1

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

$\sf Given \begin{cases} \rm{Weight \: of \: ladder \: (R) \: = \: 250 \: N} \\ \rm{ Coefficient \: of \: Friction \: (\mu) \: = \: 0.3} \end{cases}$

Here we have to find the maximum amount of frictional force between ladder and Floor .

Now, we know that,

$\large{\boxed{\sf{F \: \propto \: R}}}$

Here, Normal Reaction always increases with increase in force. So, Force is directly proportional to Normal Reaction

$\implies {\sf{F \: = \: \mu R}} \\ \\ \implies {\sf{F \: = \: 0.3 \: \times \: 250}} \\ \\ \implies {\sf{F \: = \: 75}} \\ \\ {\underline{\sf{\therefore \: Force \: of \: Friction \: is \: 75 \: N}}}$

________________________________

After Removing the Proportionality symbol there comes a constant known as coefficient of frictional force mu.

• It have no units
• It have no dimensions
• And mu = F/R

Answer 2

Answer:

Given:

Weight of ladder = 250 N

The vertical wall is smooth , whereas the horizontal lower wall is rough.

Coefficient of Friction between ladder and floor = 0.3

To find:

Max force available in-between ladder and floor.

Concept:

The Ladder and floor interface exerts force on each other. As the force by the ladder increases , the static Friction also increases.

At one time, both the forces become equal and equilibrium state is reached.

This is called Limiting friction. The is the max force upto which the ladder can stay upright.

Calculation:

$\huge{ \sf{ \green{F_{max} = \mu \times N}}}$

$\huge{ \sf{ \implies \green{F_{max} = 0.3 \times 250}}}$

$\huge{ \sf{ \implies \green{F_{max} = 75 \:N }}}$

So final answer :

$\boxed{\huge{ \sf{ \red{F_{max} = 75 \:N }}}}$