HOW CAN I SOLVE THIS QUESTION? PLEASE HELP The movement of a locomotive piston in the cylinder is limited to 0.76 m. Assume that
1 answer:
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Answer:
ω = 0.05 rad/s
Explanation:
We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,
where,
ω = angular velocity of cylinder = ?
g = required acceleration = 9.8 m/s²
r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m
Therefore,
<u>ω = 0.05 rad/s</u>
Answer:
1) 1.31 m/s2
2) 20.92 N
3) 8.53 m/s2
4) 1.76 m/s2
5) -8.53 m/s2
Explanation:
1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s
2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration
3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.
So the maximum acceleration on the block is
4)As the box slides, it is now subjected to kinetic friction, which is
So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is
28.25 / 16 = 1.76 m/s2
5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2