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svp [43]
3 years ago
15

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

ficient of static friction needed for this claim to be possible?
Physics
1 answer:
navik [9.2K]3 years ago
8 0

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

∴

\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

\mathbf{\mu_s = 0.700}}

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

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Answer:

A) mass = 3121.58 kg

B) tension = 25940.37 N

C) tension = 25940.37 N (tension on both sides will be the same)

Explanation:

Weight of elevator = 22500 N

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Since it started from rest, initial speed is zero.

Using Newton's equation of motion we have,

S = ut + 0.5at^2

S = distance covered = 6.75 m

t = time = 3 s

a = acceleration upwards

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Substituting values, we have,

6.75 = 0(3) + (0.5 x a x 3^2)

6.75 = 4.5a

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Mass of the elevator = Weight/g

Where g = acceleration due to gravity 9.81 m/s

Mass = 22500/9.81 = 2293.58 kg

From the image below we solve from

T - 22500 = ma

T - 22500 = 2293.58 x 1.5

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On the other side,

mg - T = ma

9.81m - 25940.37 = 1.5m

(9.81 - 1.5)m = 25940.37

8.31m = 25940.37

m = 3121.58 kg (mass of counter weight)

See image below

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Part a

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