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A wheelbarrow is used to carry a load of 100Kg. The load's centre of mass lies 1/3 of the way between the axis and the point where the handles are held (the distance between the axis and the handles is 1.5m), how much effort will be required to lift the load and hold it at an angle of 30 degrees from the ground?

This is what I did :

Used the equation of static equilibrium, where E = sigma, sum of

ETk = 0

ETk = (Fm sin 30)(df) - (wt)(dwt)

0 = (Fm sin30)(0.5m) - (980 N)(1.5m)

0 = Fmsin30 - 1470Nm

Fm = 5880 N

Is the method I took correct? Just looking for some insight...

Thanks

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# Homework Help: Need Confimation on Answer Process for Wheelbarrow Physics Question

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