AS Physics moments and projectiles question

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RIF A
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Ive been trying this question but cant find answers anywhere. Can some one help

A student investigates moments by suspending a 100 cm ruler from two force meters, A and B. A and B are attached to the ruler 12.0 cm from each end. Their supports are adjusted to make A and B vertical and the ruler horizontal.
Figure 1 is a simplified diagram of the experiment. Figure 1(a)The ruler is uniform and weighs 1.12 N.
Determine the reading on A.
(b) The student suggests that the forces exerted on the ruler by A and B act as a couple. Discuss whether his suggestion is correct.
(c)The student hangs a mass of weight W on the ruler between A and B, as shown in Figure2.He adjusts the supports so that A and B are again vertical and the ruler is horizontal. The mass hangs at a distance d from A.
The reading on A is 0.82 N and the reading on Bis 0.62 N. Determine W and d.
(d)A second student sets up the same apparatus as shown in Figure 2.She suspends the mass in the same position on the ruler as in question (c).She moves the supports to make A and B vertical but does not make the ruler horizontal. Discuss whether the readings on A and B taken by this student are different to those in question (c)
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0le
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You have not really stated what exactly you are struggling with, whether you just want someone to check your answers or what?

This is a moments problem so the first thing you need to do is to draw the free body force diagram. Part a) should become clearer when you do this. One assumption that you make here is that the ruler has a uniform density, so that the centre of mass is in the middle of the ruler and for this problem, it tells you where to place the weight (force) of the ruler.
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RIF A
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(Original post by 0le)
You have not really stated what exactly you are struggling with, whether you just want someone to check your answers or what?

This is a moments problem so the first thing you need to do is to draw the free body force diagram. Part a) should become clearer when you do this. One assumption that you make here is that the ruler has a uniform density, so that the centre of mass is in the middle of the ruler and for this problem, it tells you where to place the weight (force) of the ruler.
just want to know what the answers are
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0le
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(Original post by RIF A)
just want to know what the answers are
Right, so what answers did you get?
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RIF A
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(Original post by 0le)
Right, so what answers did you get?
3a) 13.33n
3b) they dont act as a couple because they are acting in opposite directions
i dont know the rest
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Nagromicous
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(Original post by RIF A)
3a) 13.33n
3b) they dont act as a couple because they are acting in opposite directions
i dont know the rest
What was your working out to get that answer for 3a?
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RIF A
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(Original post by Nagromicous)
What was your working out to get that answer for 3a?
1.12 x 12=13.44
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Nagromicous
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(Original post by RIF A)
1.12 x 12=13.44
Okay. Why did you multiply those two numbers together?
1.12 N * 12 cm = 13.44 N cm
You want a force (Newtons) not a torque (N cm)
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RIF A
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(Original post by Nagromicous)
Okay. Why did you multiply those two numbers together?
1.12 N * 12 cm = 13.44 N cm
You want a force (Newtons) not a torque (N cm)
dont know what else to do
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Nagromicous
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(Original post by RIF A)
dont know what else to do
You don't know what to do so you take the only two numbers you see and find their product? If I were you, I would be very careful with your units. You are asked to calculate what the Newton meter should read. This should be a force in Newtons. You can't just multiply a force by 12cm and say the resulting quantity is still a force in Newtons. What if you decided to multiply by 0.12m instead (the same thing)? You would get a different force which is obviously wrong.
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RIF A
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(Original post by Nagromicous)
You don't know what to do so you take the only two numbers you see and find their product? If I were you, I would be very careful with your units. You are asked to calculate what the Newton meter should read. This should be a force in Newtons. You can't just multiply a force by 12cm and say the resulting quantity is still a force in Newtons. What if you decided to multiply by 0.12m instead (the same thing)? You would get a different force which is obviously wrong.
ok
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0le
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(Original post by RIF A)
dont know what else to do
Draw a free body force diagram of the ruler. You have the ruler itself and three forces acting on the ruler - one is its weight and the other two forcemeters pulling the ruler up to prevent it from falling.

Intuitively you can think of it like this: Imagine you have a plank of wood on the floor which weighs 5N. At each end, you and a friend (who both have the same strength, height and everything else etc) pick up the plank of wood. You both do equal amounts of "effort" to lift the wood. My understanding of the question is that it is (kind of) like this example.
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Random_Thing_12
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Would the answer to 2.a be 0.56 then? As 1.12/2 = 0.56?
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RIF A
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(Original post by Random_Thing_12)
Would the answer to 2.a be 0.56 then? As 1.12/2 = 0.56?
yea , the answers are
3a)0.56n
b)Forces are equal but don’t act in opposite directions, therefore it is not correct
c)w=0.32 and d=0.14m
d)readings would be the same
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LemmingGamez
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(Original post by RIF A)
yea , the answers are
3a)0.56n
b)Forces are equal but don’t act in opposite directions, therefore it is not correct
c)w=0.32 and d=0.14m
d)readings would be the same
Mind telling me how you got d=0.14m? haven't done the topic in almost a year and I can't find a decent explanation.
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Eimmanuel
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(Original post by LemmingGamez)
Mind telling me how you got d=0.14m? haven't done the topic in almost a year and I can't find a decent explanation.
Do you know how to set up the principle of moments equation about A? And assume that you know how to compute W.

Spoiler:
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sum of clockwise moment about A = sum of anti-clockwise moment about A

 W \times d + 1.12 \times (50 - 12) = 0.62 \times (100 -12 -12)


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