How Much Weight Can the Crane Safely Lift Without Tipping Over?

[lightonahill7]

Homework Statement

The figure above is a simple model of a crane. The base has a mass of 2000kg, and the boom has a mass of 200kg. What is the maximum mass of the ball if the crane is not to tip over? Note, in order for the crane to be stable, its center of mass cannot be outside the support base.

Homework Equations

The Attempt at a Solution

 

[lightonahill7]

What do I already know?

m1 (base) = 2000 kg.
x1 = 1 m

m2 (boom) = 200 kg.
x2 = L/2 + 1m
= (6xcos70)/2 + 1 = 2.026 (I do not know if this is right?)

m3 (weight of ball) = ?
x3 = length of boom (2.052m + 1m) = 3.05 (Am I right?)

SumCMx = (m1x1 + m2x2 + m3x3)/ m1 + m2 + m3

This is where I am getting stuck. Are we to assume that the three CM should be added together to equal zero of should the base be on the left side of the equation and the boom and weight be on the right side to solve for m3?

 

[tiny-tim]

hi lightonahill7! welcome to pf!

(try using the X₂ icon just above the Reply box )

SumCMx = (m1x1 + m2x2 + m3x3)/ m1 + m2 + m3

This is where I am getting stuck. Are we to assume that the three CM should be added together to equal zero of should the base be on the left side of the equation and the boom and weight be on the right side to solve for m3?

yes, everything's fine up to there

that's the x-coordinate of the centre of mass, so you now need to find when it equals 2 (that's where it's just about to topple)

(btw, it would have been easier to measure everything either from the centre of the base or from the right-hand corner of the base (instead of the left-hand corner), wouldn't it? )

 

[lightonahill7]

I have worked the problem two different ways, the first placing the origin for the x coordinate on the left side of the base, and the second, placing the origin at the right corner of the base.

In the first problem I would set my equation equal to two, in the second I set it equal to zero. This is what I came up with:

\sumCM = (2000(1) + 200(2.026) = (3.052)M₃)/ 2000 + 200 + M₃

2(2200 + M₃) = 2000 + 405.2 + 3.05M₃

1994.8 = 1.05M₃

M₃= 1899.80 kg

Second situation

0 = 2000(-1) + 200(0.26) + 1.052M₃

0 = -1948 + 1.052M₃

M₃ = 1851.71 kg.

I know I should get the same answer regardless of my reference point as long as I put in all the measurements correctly. Please see where I may be going wrong. Thanks

 

[tiny-tim]

hi lightonahill7!

200(0.26)

erm … 0.026 !

 

[lightonahill7]

Brilliant! - thanks for finding that error.