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Ugo [173]
3 years ago
12

If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral

lel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is θ.
Physics
1 answer:
leonid [27]3 years ago
3 0

Answer:

\theta = tan^{-1}\mu

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

F_f = mgsin\theta

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

F_f = \mu N

N = mg cos\theta

so we have

F_f = \mu (mg cos\theta)

so we will have

mg sin\theta = \mu (mg cos\theta)

so now we have

tan\theta = \mu

so maximum possible angle of the inclined plane is

\theta = tan^{-1}\mu

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a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

  • There are two forces acting on the ball, one directed upward (assuming this direction as positive, along the y-axis) which is the tension on the string (lifting force), and another aimed downward, which is the attractive force due to gravity.
  • Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
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       F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)

b)

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       F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N  (2)

c)

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