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3 years ago

14

A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the centr

al plane 0.9 m above the ground and 1.7 m behind the front axle. When moving on the level at 90 km/h the brakes applied and it comes to a rest in a distance of 50 m.
Calculate the normal reactions at the front and rear wheels during the braking period and the least coefficient of friction required between the tyres and the road. (Assume g = 10 m/s2)

1 answer:

Sergeeva-Olga [200]3 years ago

4 0

Answer:

normal reaction, front: 11,175 N
normal reaction, rear: 6,825 N
minimum coefficient of friction: 0.625

Explanation:

The speed in meters per second is ...

(90 km/h)(1000 m/km)(1 h/(3600 s)) = 25 m/s

The braking acceleration can be found from ...

a = v²/(2d) = (25 m/s)²/(2×50 m) = 6.25 m/s²

Then the braking force is ...

F = Ma = (1800 kg)(6.25 m/s²) = 11,250 N

The torque on the center of gravity is ...

T = (11,250 N)(0.90 m) = 10,125 N·m

__

If we let x and y represent the normal forces on the front and rear wheels, respectively, then we have ...

x + y = (10 m/s²)(1800 kg) = 18000 . . . . . newtons

1.7x -1.3y = 10,125 . . . . . . . . . . . . . . . . . . . newton-meters

The latter equation balances the torque due to the wheel normal forces with the torque due to braking forces.

Multiplying the first equation by 1.3 and adding that to the second, we have ...

3.0x = (1.3)(18,000) + 10,125

x = 33,525/3 = 11,175 . . . . . . . . . . . newtons normal force on front tyres

y = 18000 -11175 = 6,825 . . . . . . . .newtons normal force on back tyres

The least coefficient of friction is the ratio of horizontal to vertical acceleration, 6.25/10 = 0.625.

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Explanation:

This question is incomplete, here is the complete question:

A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N and points due west. She then turns around and rides due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 3.3 N and points due east.? (a) Find the work done by the resistive force during the round trip. (b) Based on your answer to part (a), is the resistive force a conservative force? Explain.

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distance covered by the girl due east (De) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fw) = 3.1 N due west

distance covered by the girl due west (Dw) = 5.4 km = 5400 m

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work done by the resistive force due west = (Fw.cosθ) x De

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where θ is the angle between the displacement and the force. The

displacement is due west while the force is due east, hence θ = 180°

work done (W2) = (3.3 x cos 180) x 5400 = -17,820 j

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