# Need Confimation on Answer Process for Wheelbarrow Physics Question

1. Oct 29, 2005

### ruzoo

The questions is:
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

2. Oct 30, 2005

### Andrew Mason

No.
The equation for static equilibrium is:

$$\tau = 0$$

There are two torques acting here, and they are equal and opposite:

$$M_{load}\vec g \times \vec{d_{load}} = \vec F_{hand}\times \vec{d_{hand}}$$

where $M_{load} =$ 100 kg.; $d_{load} = .5 m$ and $d_{hand} = 1.5 m$

So:
$$M_{load}gd_{load}sin(60)/d_{hand} = F_{hand}$$

$$F_{hand} = 100 * 9.80 * .5 *.87 /1.5 = 284 N.$$

This assumes that the force on the handles is always applied at right angles to the handles.

AM

Last edited: Oct 30, 2005
3. Oct 31, 2005

### ruzoo

I understand, sort of....where does the sin 60 come in? Where are you getting "60" from?

4. Oct 31, 2005

### Andrew Mason

It is the cross-product:

$$M_{load}\vec g \times \vec d_{load} = M_{load}gd\sin(\theta)$$

where $\theta$ is the angle of the force from the perpendicular to the line between the fulcrum and the point of application of the force. That angle is 60 degrees if the wheelbarrow is lifted 30 degrees from the horizontal.

AM

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