Recent content by M_G

Thanx!

Thanks !

The right end of slab?

Yes, I am:smile:...but, when I solved it for Y, I found Y=0.04m...?I think it is a very small number :rolleyes: isn't it??

I'm sorry...I'm confused a little bit:frown:, I'll try to think about it again...

ok..but there will be a torque..?

Y+1.1 = (1.3*20+1.1*80)/20+80 ...?:rolleyes: I had a bad feeling that I'm proceeding wrong, very very wrong, I don't know why I can't get it?!

you mean it will not rotate? so what will happen when this person walk on it? topple on the other side, right?

got it, by applying that formula to the problem: x1=1.3 x2=1.1 X=1.1+Y (distance between the C.G & the person) Right?

sorry, I can't get what you're pointing at...

The formula is: X=(x1*w1+x2*w2)/w Where: w=w1+w2

Homework Statement A steel girder with a mass of 1000kg and a length of 10m rests on a concrete slab, with 4m overhanging the edge. How far can a 100kg man walk on the girder? The Attempt at a Solution I supposed that the right end of the slab is the fulcrum (rotating point), and...

Hi, If there is two weights on a weightless rod, the center of gravity will be at a distance X1 from W1 and X2 from W2, and the C.G will be closer to the heavier weight...right?

Homework Statement An 80 kg hiker carries a 20 kg pack. The center of gravity of the hiker is 1.1m above the ground when he is not wearing the pack. The center of gravity of the pack is 1.3m from the ground when it is worn. How far above the ground is the center of gravityof the hiker and...