If a ladder weighing $250\, N$ 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?
- 75 N
- 50 N
- 35 N
- 25 N
The Correct Option is A
Solution and Explanation
$ R $ is the reaction of the floor on the ladder, $ w $ the weight of ladder
Also $ R=w=250\,N $ .
Given, $ \mu =0.3 $
$ \therefore F=0.3\times 250=75\,N $ .
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Concepts Used:
Equilibrium Constant
The equilibrium constant may be defined as the ratio between the product of the molar concentrations of the products to that of the product of the molar concentrations of the reactants with each concentration term raised to a power equal to the stoichiometric coefficient in the balanced chemical reaction.
The equilibrium constant at a given temperature is the ratio of the rate constant of forwarding and backward reactions.
Equilibrium Constant Formula:
Kequ = kf/kb = [C]c [D]d/[A]a [B]b = Kc
where Kc, indicates the equilibrium constant measured in moles per litre.
For reactions involving gases: The equilibrium constant formula, in terms of partial pressure will be:
Kequ = kf/kb = [[pC]c [pD]d]/[[pA]a [pB]b] = Kp
Where Kp indicates the equilibrium constant formula in terms of partial pressures.
- Larger Kc/Kp values indicate higher product formation and higher percentage conversion.
- Lower Kc/Kp values indicate lower product formation and lower percentage conversion.
Medium Kc/Kp values indicate optimum product formation.
Units of Equilibrium Constant:
The equilibrium constant is the ratio of the concentrations raised to the stoichiometric coefficients. Therefore, the unit of the equilibrium constant = [Mole L-1]△n.
where, ∆n = sum of stoichiometric coefficients of products – a sum of stoichiometric coefficients of reactants.