What is the maximum mass of rocks one can lift using a wheelbarrow PLEASE HELP

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SUMMARY

The maximum mass of rocks that can be lifted using a wheelbarrow is determined by analyzing the forces and torques involved. Given a vertical force of 500N exerted by the man, the center of mass of the rocks positioned 0.25m from the wheel, and a handle length of 1.5m at a 35-degree angle, calculations reveal that the maximum mass is approximately 533.2 kg. This is derived from the torque equilibrium equation and the application of Newton's laws. The correct approach involves balancing the torques and calculating the resultant forces accurately.

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mhayer7
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What is the maximum mass of rocks one can lift using a wheelbarrow PLEASE HELP!

Homework Statement


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?

Homework Equations



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

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!
 
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would this helps?

you should also put in mind the angle
 

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

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
 


hey drizzle,


so using your formula,

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
 


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


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?
 


no, skip this last calculations you’ve done, what you’ve done previously was good so far [that is, with having 3000N as the force applied on the right side of the wheelbarrow -using Newton’s 3rd law-], now you should calculate the sum of vertical forces [this should equal to zero], then you’ll have the mass of the rock.
 

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