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

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

Hi Alex,

Drawing the diagram will be a bit trickey since there is only text available to me, so I will try my best to describe it. The rule is modeled as a uniform rectangle of length 1m. Assuming that the weight is put on the left of the rule (it does not matter since the rule is symmetric and uniform) then there is a downward force of 4N due to the weight at the left end of the rule. There is also an upwards force (usually tension) of unknown magnitude acting upwards 0.1m from the left end of the rule that is due to the suspension. Finally there is the weight of the rule acting downwards through the centre of the rule due to the mass of the rule and acts here because it is the centre of mass of the rule. This weight is 0.5m from the left of the rule and 0.4m to the right of the suspension as it acts at the centre of the 1m rule. Since we do not know its weight but wish to calculate it we will call it W for now.

To calculate the weight we can take moments about the suspension since the rod is in static equilibrium (not rotating). This is a good choice since we do not need to calculate the force due to the suspension. The anti-clockwise moment (moment which will cause anti-clockwise rotation if it is the only force) about this point is due to the weight at the left end of the rod and has a magnitude of 0.1X4=0.4Nm (the moment of the force about a point is equal to the product of its magnitude and perpendicular distance from the point). The clockwide moment is due to the weight of the rod and has a magnitude of 0.4xW. Since the rod is not rotating the sum of the anti-clockwise moments is equal to the sum of the clockwise moments which means we may equate these tow moments:

0.4=0.4W or W=0.4/0/4=1N

The weight of the rod is then one newton. This can be checked by calculating the force due to the suspension by taking moments about the centre of mass and then taking moments about the left end to see if the result is again W=1N.

Jez

Answer 2

a) The main part of the diagram is a staright line, which represents the rule. Draw it about 10cm long. Next draw the weight acting as a downwards force at one end of the rule. Think about what other forces there are: the weight of the rule going downwards, and the reaction at the point of suspension. The weight of the rule acts a half the length of the rule, i.e. 50cm from either end.

b) For balance to happen, the clockwise moment must equal the anticlockwise moment. If you take moments about the point of suspension, then the reaction force has no moment, and so you only have one unknown: the weight of the rule. The moment from the 4N weight is 4N*0.1m=0.4Nm. The moment from the weight (W) is W*0.4m=0.4WNm. The moments are equal, so 0.4W=0.4, so W=1N.

c)Not asked, but it might be on other questions. The overall vertical force must equal zero, so the reaction force R=4+W=4+1=5N

Answer 3

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 4

To solve this one, we have to compare moments. I assume the rule has a constant cross-section and composition (give up now if it doesn't). Thus the weight of 0.9m of rule, acting 0.45m from the balance point, exactly matches the weight of 0.1m of rule, acting 0.05m from the fulcrum, plus 4N acting 0.1m from the fulcrum.

So let's call the weight of the rule "R" (for Rule!!)
0.9R*0.45=0.1R*0.05+4*0.1 (all moments in Newton-metres or Nm)

Thus (0.405R)Nm = (0.005R)Nm+0.4Nm
Subtract 0.005R from both sides:
0.4R = 0.4Nm
So R = 1
so your rule weighs 1 Newton.

Can't draw on this site, sorry, but the key your diagram is that the for a uniform rod, the resultant of each side of the fulcrum acts midway along the length.

Answer 5

A/B = C/D

Where:
A = Mass added (4N)
B = Old total mass (mass of your rule)
C = Change in CofG position (0.4m)
D = Distance from weight to new CofG (0.1m)

You can assume that your rule is uniform and therefore weight is evenly distributed and CofG starts at 0.5m (center of rule) into the rule.

therefore: 4n/B = 0.4/0.1 and B = 4/4 = 1n

draw something that looks like this:

4N
------------------------------------------
....^
Good luck.