# force problem

• Mar 5th 2009, 07:17 AM
bolton78
force problem
First of all,hi to everybody
newbie here

Anyway heres my problem,i need to lift a 70kg weight using a 100mm bore cylinder,see attached word doc
Thanks
• Mar 5th 2009, 07:56 AM
HallsofIvy
One arm of 325 mm. It's not clear where the 70 kg mass is but I will assume it is at the end of the 325 mm arm. The other arm, at which the force is applied, is 100 mm long. 70 kg mass is a weight of 70g= 70(9.81)= 686.7 N. If the lever were to move through an angle of $\displaystyle \theta$ radians, the mass would swing through a distance $\displaystyle 325\theta$, while the force, F, is applied over a distance of $\displaystyle 100\theta$. Since "work= force*distance" and work (energy) is conserved, we must have $\displaystyle (325)(686.7)(\theta)= F(100)(\theta)$. Now the $\displaystyle \theta$s cancel and we have F= (325)(686.7)/100= 2231 N. Notice that 2231/686.7= 3.25= 325/100, the ratio of the lengths of the arms.
• Mar 5th 2009, 12:09 PM
dun.nu.nu.nu.nu.nu.Piman
well you just need to use moments
moment =force x perpendicular distance from the pivote

so the mass is the force that you are given which is 75g(in newtons g= 9.81)
which acts in the centre which is 212.5mm from the rigtwhich is 112.5mm (convert it into metres so it becomes 0.1125 m)from the pivote.
then you need to know that all clockwise moments= the sum of the anticlockwise moments.
so 0.1125 x 75g = 0.1 x y(the force you want)
so just solve
0.1125 x 75 x 9.8
0.1 = y

82.6875
0.1 =y

y= 826.875

hope it helps
• Mar 5th 2009, 02:02 PM
David Green
Quote:

Originally Posted by bolton78
First of all,hi to everybody
newbie here

Anyway heres my problem,i need to lift a 70kg weight using a 100mm bore cylinder,see attached word doc
Thanks

I think you need to read up on Moments and Forces. The following information is only a guide to the type of information you require.

F x 100 = (100 x 70) + (325 x 70)
F x 100 = 7000 + 22750
F x 100 = 29750
F = 298

Hope this helps(Hi)