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spin [16.1K]
2 years ago
10

A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it

rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?
Physics
1 answer:
vredina [299]2 years ago
3 0

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

F = ma...(I)

Using formula of friction force

F= \mu m g....(II)

Put the value of F in the equation (II) from equation (I)

ma=\mu mg....(III)

a = \mu g

Put the value in the equation (III)

a=0.24\times9.8

a=2.352\ m/s^2

We need to calculate the distance,

Using equation of motion

s = ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}2.352\times(3.0)^2

s=10.584\ m\ approx\ 11\ m

Hence, The distance is 11 m.

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A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity
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γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches  = 0.0508 m

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H_{nom}=2hp=1491.4W

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

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w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m

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b)

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dip = \frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm

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