Please someone give me an answer: A uniform metre rule has a 4N weight hanging from one end. The rule balanc?

Question

Please someone give me an answer: A uniform metre rule has a 4N weight hanging from one end. The rule balances when suspended from a point 0.1 from that end.

a) draw a diagram to show the rule and the forces on it.
b) calculate the weight of the rule

please show workings out and how to draw it. Thanks

Answer 1

a)
Draw a metre rule and at the 10cm mark, place a force pointing upwards. At the 0cm mark, place the 4N weight. The balancing force will be the weight of the uniform metre rule. Place this weight, W, at the 50cm mark as the ruler is uniform.

b) Using the conservation of moments,

Clockwise moments = Anticlockwise moments
4 x 10 = W x (50 - 10)
40 = 40W
W = 1 Newton

(Note: Moments = Force x distance from the object to the fulcrum or pivot)

Answer 2

Draw a line representing the meter rule. Mark a point 10% of the way from one end, dividing the rule into two sections. Your force vectors are the weight at one end, and the weight of the rule's two sections from the center of gravity of each section. Since the rule is uniform, the center of gravity is at the center of each section. (Hint: the center of gravity of the .1 section is at .05 from the end, just midway to the mark.) The rule's weight for that section is simply the proportion of the full weight represented by that section (the weight of the .1 section is just .1 times the full weight). You will have three force vectors, two on the short section and one on the long section. The rule is balanced, so there is no torque on the rule. This means the torque from the two forces on the small section (force times distance from the mark for each force) must equal the torque from the large section. This allows you to solve the problem.

Answer 3

Weight of ruler = (x)
Moment of inertia from weight = 0.1x4N
=0.4N
A distributed mass can be considered as a point weight acting from the center of gravity, thus:
Moment of inertia from ruler on same side as weight =0.1(x)/2
Moment of inertia from ruler on other side =0.9(x)/2
0.45(x)=0.05(x)+0.4
(x)=1

....0.1x.......................0.9x
___I________________I______________
I...........^
4N

<-0.1m><----------------0.9m----------------->

Not easy to draw in ascii. Vertical bars should be downward arrows.
Note the point weights are 1/2 way from the fulcrum to the end.

Answer 4

Do it yourself, I didn't have Yahoo Answers when I was at school