Question

The dimensional formula for frictional force is

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Answer 1

The dimensional formula for frictional force is [M0 L0 T0].

Derivation of the formula:

• Frictional Force [Normal Force]-1 = Coefficient of Friction ()... (1)

• Since, Force (F) = Mass × acceleration = Mass × velocity × [Time]-1

• And, velocity = Displacement × [Time]-1

• The dimensions of Force = [M] × [LT-1] × [T]-1 = [M1 L1 T-2] . . . . (2)

• On substituting equation (2) in equation (1) we get,

• Coefficient of friction (μ) = Frictional Force × [Normal Force]-1

• Or, μ = [M1 L1 T-2] × [M1 L1 T-2]-1 = [M0 L0 T0].

• Therefore, the coefficient of friction is dimensionally represented as [M0 L0 T0].

The complete question is ''Which of the following is the dimension of the coefficient of friction? (a) [M2L2T], (b) [M0L0T0], (c) [ML2T-2], (d) [M2L2T-2].

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Answer 2

Answer:

The dimensional formula for frictional force is $[M^1L^1T^{-2}]$.

Explanation:

Frictional force is given by

$f=\mu N$ ...... (1)

where $\mu$ = coefficient of friction and $N$ = normal force.

Since coefficient of friction is unitless and dimensionless.

⇒ $\mu= {[M^0L^0T^0]}$

Also, normal force = mass × acceleration

⇒ $N=m$ × $a$

⇒ $N=m$ × $\frac{v}{t}$  (Since acceleration = velocity/time)

⇒ $N=m$ × $\frac{x/t}{t}$  (Since velocity = displacement/time)

⇒ $N=m$ × $\frac{x}{t^2}$

Thus, dimension of normal force is,

$N=[M^1L^1T^{-2}]$

Substitute the dimensions $[M^1L^1T^{-2}]$ for $N$ and ${[M^0L^0T^0]}$ for $\mu$ in the equation (1) as follows:

$f= {[M^0L^0T^0]}[M^1L^1T^{-2}]$

⇒ $f= [M^1L^1T^{-2}]$

Therefore, the dimensional formula for frictional force is $[M^1L^1T^{-2}]$.

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