1. Jun 5, 2009

### mhayer7

1. The problem statement, all variables and given/known data
A man is using a wheelbarrow to transport rocks across the garden. The man can generate a maximal vertical force of 500N to lift the wheelbarrow. If the centre of mass of the rocks is .25m horizontally from the wheel, and the handle of the wheelbarrow is 1.5m at an angle of 35 degress from the horizontal, what is the maximum mass of rocks he can lift?

2. Relevant equations

sum or torques equation. f=ma, phythagorean theorem.

3. The attempt at a solution

I tried drawing a diagram for this but ended up confusing myself. I know i need to use similar triangles to try and find the max weight since we have the force applied by the man but am not sure how to do it. I know it will be something like MASS OF ROCKS TIMES GRAVITY X .25M = 500N TIMES 1.5M. But I am very confused as to how to accomplish this. Any help is appreciated!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 6, 2009

### drizzle

would this helps?

you should also put in mind the angle

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3. Jun 7, 2009

### drizzle

hey mhayer7, you don't have to PM me, just post here and you'll get the help from me or others

did you try this;
sum of torques=0
Fmax[1.5]sin35o=F[0.25]sin35o

note that F here is the force applied on the wheel

4. Jun 7, 2009

### mhayer7

hey drizzle,

Fmax[1.5]sin35o=F[0.25]sin35o

we wud end up with 500 which is the force applied by the man

500[1.5]sin35o/.25[sin35]= 3000 N

3000N / 9.81 = 305.8 kg's

i hope that's right

5. Jun 7, 2009

### drizzle

the 3000N you got isn't the weight of the rock!

6. Jun 8, 2009

### mhayer7

sorry so if i did it like this

since i have 500 and sin 35 using that i can get the perpindicular force which is also the hypoteneuse . so sin35 = 500/z . z = 871.7 then plugging that into your equation

we get 871times 1.5 times sin35 = .25sin35 F

solving for F we get 5230.3
dividing that by 9.81 we get 533.2 kg

Is that right?

7. Jun 8, 2009