Posted by **Andrea** on Wednesday, November 7, 2007 at 5:06pm.

A lever is 5m long. The distance from fulcrum to the weight to be lifted is 1m. If the worker pushes on the opposite end with 400N, what is the max weight that can be lifted?

l=5m

h=1m

F=400N

F_g=?

torque=F_g*l

I first must find torque so that I can plug it into the above equation.

F=torque/r

400N=torque/5m

=2000Nm

thus

200Nm =F_g(5m)

F_g=400N

is this correct? Or would r=1m? Does the 1m in the question matter for this problem? Or is it just extra information?

- Physics -
**bobpursley**, Wednesday, November 7, 2007 at 5:24pm
No. The lever distance is from the point of force to the fulcrum. Here, it is 4 m (not five meters as you did).

400*4=torque= W*1

W= 1600N

- Physics -
**Joy**, Thursday, February 17, 2011 at 9:58am
If you want to make it more simple !!, here's another answer.

By newtons first law, the sum of all static forces at a balanced system is zero. Otherwise it will be moving...!!!

Add all vertical forces:

Here you have to consider a hidden force R form Reaction at support in opposite direction of forces (here at fulcrum acting upwards)

W + 400 - R = 0

R = W+400

Add all Torques about any point within or outside the lever. Better choose the point of unknown force or fulcrum to make it simple, as in the previous answer. Distances are based on the point you select.

Let's take about point of unknown force W:

W*0 + 400*5 - R*1 = 0 ;

R = 400*5

plug in value of R from previous step:

(W+400) = 400*5 ;

W = 400*5 - 400 = 2000-400 = 1600 N

quite simple OK. Now try with any other point.

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