Question

A 1500-kg car enters a section of curved road in the horizontal plane
and slows down at a uniform rate from a speed of 100 km/h at A to a
speed of 50 km/h as it passes C. The radius of curvature ρ of the road at
A is 400 m and at C is 80 m. Determine the total horizontal force exerted
by the road on the tires at positions A, B, and C. Point B is the inflection
point where the curvature changes direction.

Answer 1

Answer-

Incomplete question check attachment for diagram

Explanation: Given that, Mass of car M = 1500kg

Enter curve at Point A with speed of

$Va = 100km/hr = 100\times 1000/3600$

$Va = 27.78m/s$

The car was slow down at a constant rate till it gets to point C at  speed of

$Vc = 50km/r = 50\times 1000/3600$

$Vc = 13.89m/s$

Radius of curvature at point A

$p = 400m$

Radius of curvature at point B

$p = 80m$

The distance from point A to point B as given in the attachment is $S=200m$

We want to find the total horizontal  forces at point A, B and C exerted by the road on the tire The constant tangential acceleration can be calculated using the equation of motion

$Vc^2= Va^2 + 2as$

$13.89^2 = 27.78^2 + 2 \times a \times 200$

$192.9 = 771.6 + 400a$

$400a = 192.9-771.6$

$400a = -578.7$

$a = -578.7 / 400$

$a = -1.45 m/s^2$

$at = -1.45m/s^2$

The tangential acceleration is $-1.45m/s^2$ and it is negative because the car was decelerating

Since the car is slowing down at a constant rate, the tangential acceleration is equal at every point

At point A

$at = -1.45m/s^2$

At point B

$at = -1.45m/s^2$

At point C

$at = -1.45m/s^2$

Now,

We can calculate the normal component of acceleration(centripetal acceleration) at each point since we know the radius of curvature

The centripetal acceleration is calculated using

$ac = v^2/ p$

At point $A ( p = 400)$

$an = Va^2/p = 27.78^2/ 400$

$an = 1.93 m/s^2$

At point B (p = ∞), since point B is point of inflection

Then,

$an = Vb^2/p = Vb/$ ∝$=0$

an = 0

At point C $( p = 80m)$

$an = Vc^2/p = 13.89^2/ 80 = 2.41m/s^2$

$an = 2.41 m/s^2$

Then,

The tangential force is

$Ft = M. at\\Ft = 1500 \times 1.45\\Ft = 2175 N.$

Since tangential acceleration is constant, then, this is the tangential force at each point A, B and C

Now, normal force

Point A ( an = 1.93m/s²)

Fn = M•an

Fn = 1500 × 1.93

Fn = 2895 N

At point B (an=0)

Fn = M•an

Fn = 0 N

At point C (an= 2.41m/s²)

Fn = M•an

Fn = 1500 × 2.41

Fn = 3615 N

Then, the horizontal force acting at each point is Using Vector of right angle triangle

F = √(Fn² + Ft²)

At point A

Fa = √(2895² + 2175²)

Fa = √13,111,650

Fa = 3621 N

At point B

Fb = √(0² + 2175²)

Fb = √2175²

Fb = 2175 N

At point C

Fc = √(3615² + 2175²)

Fc = √17,798,850

Fc = 4218.88 N

Fc ≈ 4219N

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