A wheelbarrow has its center of mass located in the center of the load its carri

Question

A wheelbarrow has its center of mass located in the center of the load its carring. The task is to compare two different ways of moving the load; one with force F1                applied by lifting and pushing forward, the other force F2 applied by lifting and pulling the wheelbarrow behind you. The length of the wooden handles is 2 meters,                and the mass of the load plus the wheelbarrow is 80kg. The axle is 30cm from the front end, and F1 is applied 10 cm from the other end of the handles. The handles                make an angle of 30

Explanation / Answer

1)
theta = 30
phi = 50

d = 0.10 m

p = 1 m

s = 0.30 m

x direction:

2 F1 cos(50) = f

y direction:

2 F1 sin(50) + N = m g

torque about the axle:

torque = force x distance x sin(theta)

therefore:

(2 F1) r1 sin(theta1) = m g r2 sin(theta2)

(2 F1) (2-d-s) sin(phi+theta) = m g (1-s) sin(theta)

(2 F1) (2-0.10-0.30) sin(50+30) = m g (1-s) sin(30)

==> F1 = 0.1111 m g

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2)

F1 = 0.1111 m g

==> F1 = 0.1111 m g = 0.1111*80*9.8 = 87.1 N

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3)

x direction:

2 F2 cos(40) = f

y direction:

2 F2 sin(40) + N = m g

torque about the axle:

torque = force x distance x sin(theta)

therefore:

(2 F2) r1 sin(theta1) = m g r2 sin(theta2)

(2 F2) (2-d-s) sin(theta-phi) = m g (1-s) sin(theta)

(2 F2) (2-0.10-0.30) sin(40-30) = m g (1-0.3) sin(30)

==> F2 = 0.6299 m g

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4)

F2 = m g/2.06 = 0.6299*80*9.8 = 494 N

f = 2 F2 cos(40) = 2*494*cos40 = 757 N

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