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shusha [124]
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)​

Physics
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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The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t
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Now if the same formula is used for revolution of Earth around the sun

T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}

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4 0
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For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

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Moment with respect to ELBOW

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        i = ⅓ 2.3 0.47² + 0.5 0.47²

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        Στ = 0.2798 10

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5 0
3 years ago
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