Part 1 The average truck today has a mass of 3.0 x 10^2 kg, and when acceleratin
ID: 1639370 • Letter: P
Question
Part 1
The average truck today has a mass of 3.0 x 10^2 kg, and when accelerating from rest, covers 0.25 miles in 17 seconds. Each rim and tire together has a diameter of 86 cm and a mass of 12.1 kg. If we agree the rim and tire have the shape of a solid disk that rotates through its geometric center, what would be the moment of inertia, in kg-m^2 of each tire?
Part 2
The wheels, axle, and handles of a wheelbarrow weigh 60.0 N. The center of gravity of the wheelbarrow alone is 0.600 m from the axel. The distance from the axle to the hands of the operator is 1.30 m. The load chamber and contents weigh 525 N and is positioned 0.750 meters from the axle. What force, in Newton’s, is divided between the operator’s hands?
Explanation / Answer
Part1.
Radius of tire,R = 86/2 = 43 cm = 0.43 m
Moment of inertia of each tire(solid disk), I = (1/2)MR2 = 0.5*12.1*0.432= 1.12 kg-m2
Part2.
Let the force F is acting at the operator hands.
Taking moment about the axle,
F*1.3 = (60*0.6) + (525*0.75)
solving, F = 330.6 N
Force on each hand, f = F/2 = 165.3 N