Exercise 7.3.06 Consider the two-wheel resistance. The coefficient of rolling re

Question

Exercise 7.3.06 Consider the two-wheel resistance. The coefficient of rolling resistance cart shown. Determine the force (both horizontal and vertical components) that must be applied to the handle of the cart to pull it at a constant speed due to rolling is 0.00025. The cart is equipped with 600-mm diameter high-pressure tires that fit loosely over a 25-mm diameter fixed axie, the cart weighs 250N and is carrying 1080 N of yard waste; and the center of mass of the cart and waste is 65 mm in front of the axle. ignore bearing friction Assume L-340 mm, 14=490 mm, L,=410 mm. Hi =300 mm, H2 =450 mm.

Explanation / Answer

given, coefficient of rolling resistance k = 0.00025
weight of cart, W = 250 N
weight of load, w = 1080 N
distance of the center of mass form the axle, d = 65mm

let tehe force be F applied at angle theta to the handle
then from moment balance about the axle
Fsin(theta)*(L1 + L2) = (W + w)d
L1 = 340 mm
L2 = 490 mm
Fsin(theta) = 104.1566

from force in vertial direction
Fsin(theta) + N = (W + w)
N = 1225.8434 ( Normal reaction force)
heence rolling resistance f = kN
hence
Fcos(theta) = 0.00025*N = 0.3064596 N

hence horizontal compoennt of force = 0.3064596 N
vertical component = 104.1566 N

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