Question

A turn of radius 20 m is banked for the vehicle of mass 200 kg going at a speed of 10 m//s. Find the direction and magnitude of frictional force (a) 5 m//s (b) 15 m//s Assume that friction is sufficient to prevent slipping. (g=10 m//s^(2))

Answer 1

Given :

The mass of vehicle = m = 200 kg

The radius = r = 20 meters

The speed of vehicle = v = 10 m/s

g = 10 m/s²

To find :

The direction and magnitude of frictional force

Solution :

From figure ,

Let The force acting = F Newton

The force acting along x-axis = $F_x$ = $\dfrac{mv^{2} }{r}$

The force acting along y-axis = $F_y$ = 0

Now,

N sin θ + F cos θ = $\dfrac{mv^{2} }{r}$              ..........1

And,

N cos θ − F sin θ = mg                .............2

here   θ  = $tan^{-1}$ ( $\dfrac{1}{2}$ )

( a )  

For velocity = v = 5 m/s

Solving eq 1 and eq 2

Cos θ (N sin θ + F cos θ ) - Sinθ (  N cos θ − F sin θ ) = Cos θ× $\dfrac{mv^{2} }{r}$ - Sin θ ×  mg

Or,  ( N Cos θ Sin θ - N Cos θ Sin θ ) + ( F Cos²θ + F Sin²θ) = Cos θ× $\dfrac{mv^{2} }{r}$ - Sin θ ×  mg

Or,  0 + F ( Cos²θ +  Sin²θ) = Cos θ× $\dfrac{mv^{2} }{r}$ - Sin θ ×  mg

Or,      F = Cos θ× $\dfrac{mv^{2} }{r}$ - Sin θ ×  mg

∵        θ  = $tan^{-1}$ ( $\dfrac{1}{2}$ ) = 26.5

So,  F =  Cos 26.5 × $\dfrac{200\times (5)^{2}}{20}$ - Sin 26.5 × 200 × 10

Or,   F = 223.5 - 892

∴   Force = $\left | F \right |$ = 668.5  Newton      ( downward )

( b )

For velocity = v = 15 m/s

Solving eq 1 and eq 2

Cos θ (N sin θ + F cos θ ) - Sinθ (  N cos θ − F sin θ ) = Cos θ× $\dfrac{mv^{2} }{r}$ - Sin θ ×  mg

Or,  ( N Cos θ Sin θ - N Cos θ Sin θ ) + ( F Cos²θ + F Sin²θ) = Cos θ× $\dfrac{mv^{2} }{r}$ - Sin θ ×  mg

Or,  0 + F ( Cos²θ +  Sin²θ) = Cos θ× $\dfrac{mv^{2} }{r}$ - Sin θ ×  mg

Or,      F = Cos θ× $\dfrac{mv^{2} }{r}$ - Sin θ ×  mg

∵        θ  = $tan^{-1}$ ( $\dfrac{1}{2}$ ) = 26.5

So,  F =  Cos 26.5 × $\dfrac{200\times (15)^{2}}{20}$ - Sin 26.5 × 200 × 10

Or,   F = 2011.5 - 892

∴    Force = $\left | F \right |$ = 1119.5  Newton      ( downward )

Hence,

( a ) The direction and magnitude of frictional force is 668.5 Newton

( b )  The direction and magnitude of frictional force is 1119.5 Newton

Answer