Uniform plank resting on two supports.

[annanause]

Homework Statement

A uniform plank, of weight 30N and length 3m, resting on two supports. The supports are 0.5m and 2.0m from the left-hand end of the plank. A weight of 18N is suspended from the left-hand end of the plank.

By how much should the weight on the left-hand en be increased so that the reaction Y becomes zero?

Homework Equations

The reactions that have been calculated from the first part of the assignment are 107N for point X and 196N for point Y .

The Attempt at a Solution

Calculations from part a) of the question, which was about finding the reactions at X and Y:
X x 1 = Y x 0.5
mg = 30N x 9.8m/s
X = mg - Y
(mg - Y) x 1 = Y x 0.5
1 x mg - 1 x Y = 0.5 x Y
1 x mg = 1.5 x Y
1 x mg / 1.5 = 1 x 30 x 9.8 / 1.5 = 196N

X + Y = X + 196N = mg
X = mg - 196 = 30 x 9.8 - 196 = 98N
98 + (18 x 0.5) = 107N

X = 107N
Y = 196N

How do I rearrange the calculations from calculating the reactions into finding how much weight that needs to be added? I am pretty sure I need to consider the length (d) of the plank - I can't simply just add 89N to make the 107N into 196N, right?

If there is a general formula/calculation for this, that would be very helpful.

 

[BvU]

Hi an,

I can't simply just add 89N to make the 107N into 196N, right?

Indeed, no. That would change the load on the other guy as well.

If there is a general formula/calculation for this that would be very helpful

There is. Should be in your list of relevant equations: force balance and moment balance.
Please show your working for the first part you solved. Hopefully in terms of the variables (i.e the relationships, not the numerical values). Then consider what change is needed for part 2.

 

[annanause]

Hi an,

Indeed, no. That would change the load on the other guy as well.
There is. Should be in your list of relevant equations: force balance and moment balance.
Please show your working for the first part you solved. Hopefully in terms of the variables (i.e the relationships, not the numerical values). Then consider what change is needed for part 2.

I added my calculations from the first part of the question, but need help in understanding the second part. I have been searching my books and the internet and can't find a formula to calculate the second part of this question.

 

[haruspex]

mg = 30N x 9.8m/s

The plank has a weight of 30N, not a mass of 30kg.

 

[annanause]

The plank has a weight of 30N, not a mass of 30kg.

So I don't need to multiply with 9.8? I can just keep it 30N ?

 

[annanause]

So I don't need to multiply with 9.8? I can just keep it 30N ?

So mg = 30N ?

 

[haruspex]

So I don't need to multiply with 9.8? I can just keep it 30N ?

Yes.

 

[annanause]

Yes.

Thank you. Do you have any insight on how to calculate how much weight should be added to X to make Y zero?

 

[annanause]

Yes.

The answer for X is a negative number when I leave out x9.8, could that be correct?

The new answers are:
X = -157N
Y = 20N

 

[haruspex]

The answer for X is a negative number when I leave out x9.8, could that be correct?

The new answers are:
X = -157N
Y = 20N

Please explain how you get this equation:

X x 1 = Y x 0.5

 

[annanause]

Please explain how you get this equation:

The length from the center of the plank to support X is 1m and to Y is 0.5m, so the force at point X multiplied by the length should be equal to the force multiplied by the length at point Y

 

[haruspex]

The length from the center of the plank to support X is 1m and to Y is 0.5m, so the force at point X multiplied by the length should be equal to the force multiplied by the length at point Y

Are there no other forces on the plank which have a moment about the plank's centre?

 

[annanause]

Are there no other forces on the plank which have a moment about the plank's centre?

There is a weight of 18N on the left hand end of the plank, which I added on after calculating the force of the plank only on the supports.

Here is the full first question: A uniform plank, of weight 30N and length 3m, resting on two supports. The supports are 0.5m and 2.0m from the left hand end of the plank. A weight of 18N is suspended from the left hand end of the plank.

a) Find the reactions, X and Y, at the two supports.

 

[haruspex]

There is a weight of 18N on the left hand end of the plank, which I added on after calculating the force of the plank only on the supports

That's no excuse for leaving it out of the moments equation.

 

[annanause]

That's no excuse for leaving it out of the moments equation.

So how would that work on support Y? Is it 2m away? Then I have two separate weights with different distances to the two supports, so it will be two separate equations?

 

[annanause]

That's no excuse for leaving it out of the moments equation.

The new equation will be:
X - 30N x 1m = 30Nm
18N x 0.5m = 9Nm

Y - 30N x 0.5m = 15Nm
18N x 36Nm

X = 39Nm
Y = 51Nm

Somehow it does not add up that there is more force applied on support Y, that is the furthest away from the extra weight...

 

[haruspex]

X - 30N x 1m = 30Nm

How do you get that? I thought you were taking moments about the mid point of the plank.

 

[annanause]

How do you get that? I thought you were taking moments about the mid point of the plank.

It should say X = 30N x 1m = 30N
Support X is 1m away from the midpoint

 

[haruspex]

It should say X = 30N x 1m = 30N
Support X is 1m away from the midpoint

There are four vertical forces acting on the beam. If you take moments about an arbitrary point, all four forces will have a moment about that point, so all four should feature in the equation. You can make things a bit simpler by picking the point of action of one of the forces as the axis, so that force has no moment, but the other three forces will still be in the equation. I do not understand why you keep writing moments equations that only include two forces.

 

[annanause]

There are four vertical forces acting on the beam. If you take moments about an arbitrary point, all four forces will have a moment about that point, so all four should feature in the equation. You can make things a bit simpler by picking the point of action of one of the forces as the axis, so that force has no moment, but the other three forces will still be in the equation. I do not understand why you keep writing moments equations that only include two forces.

It is because our teacher didn't go through any of this with us and I have no clue what I am doing. I don't even know what and arbitrary point is or how to pick a point :(

 

[haruspex]

It is because our teacher didn't go through any of this with us and I have no clue what I am doing. I don't even know what and arbitrary point is or how to pick a point :(

Ok.

An arbitrary point means a point that is not special in any way. E.g. It could be the point 13cm from the left end. But obviously it is better to pick a point like the centre of the plank, so let's go with that. (The left hand end, or either support would work as well.)
The weight of the plank acts through our chosen axis, so has no moment about it (displacement is zero).
The weight at the left and the support at the right each exert an anticlockwise torque anout the axis, while th other support exerts a clockwise torque about it. The sum of the torques must balance. Can you write that as an equation?

 

[annanause]

Ok.

An arbitrary point means a point that is not special in any way. E.g. It could be the point 13cm from the left end. But obviously it is better to pick a point like the centre of the plank, so let's go with that. (The left hand end, or either support would work as well.)
The weight of the plank acts through our chosen axis, so has no moment about it (displacement is zero).
The weight at the left and the support at the right each exert an anticlockwise torque anout the axis, while th other support exerts a clockwise torque about it. The sum of the torques must balance. Can you write that as an equation?

I will look into this and make an equation for it tonight, have to get to school now. Thank you for helping out!

 

[annanause]

Ok.

An arbitrary point means a point that is not special in any way. E.g. It could be the point 13cm from the left end. But obviously it is better to pick a point like the centre of the plank, so let's go with that. (The left hand end, or either support would work as well.)
The weight of the plank acts through our chosen axis, so has no moment about it (displacement is zero).
The weight at the left and the support at the right each exert an anticlockwise torque anout the axis, while th other support exerts a clockwise torque about it. The sum of the torques must balance. Can you write that as an equation?

I have now made some calculations to solve both of the parts of this question.

a) Anti clockwise: F x 0.5
Clockwise: Y x 2.5
Anti clockwise = Clockwise
0.5 x F = 2.5 x Y
0.5 x 18 = 2.5 x Y

Forces against gravity: X + Y
Gravity forces: Fg + F
Forces against gravity = Gravity forces
X + Y = Fg + F
X + Y = 30N + 18N

9 = 2.5Y -> Y = 9/2.5 = 3.6N
X + Y = 48N -> X + 3.6N = 48N -> 48N - 3.6N = 44.4N

Y = 3.6N
X = 44.4N

b) Balance between moments:
F x 0.5m and Fg x 1.0m
F x 0.5m = 30 x 1.0m -> F = 30/0.5 = 60N

 

[haruspex]

a) Anti clockwise: F x 0.5
Clockwise: Y x 2.5

Which point are you taking moments about to get that? Yet again, you only have two forces in the equation. Depending on the axis, there will be either three or four.

If we take moments about the left end:
18N x 0 = 0
X x 0.5 acw
30N x 1.5 cw
Y x 2 acw
Or about the X point:
18N x 0.5 acw
X x 0 = 0
30N x 1 cw
Y x 1.5 acw
Etc.

See if you can write out the three moments when taking the mid point of the plank as the axis.

 

[annanause]

Which point are you taking moments about to get that? Yet again, you only have two forces in the equation. Depending on the axis, there will be either three or four.

If we take moments about the left end:
18N x 0 = 0
X x 0.5 acw
30N x 1.5 cw
Y x 2 acw
Or about the X point:
18N x 0.5 acw
X x 0 = 0
30N x 1 cw
Y x 1.5 acw
Etc.

See if you can write out the three moments when taking the mid point of the plank as the axis.

I included the X x 0 in the equation so it had 3 forces, thank you for the help :)

 

[haruspex]

I included the X x 0 in the equation so it had 3 forces, thank you for the help :)

No, there must be three nonzero moments. As I wrote, there are four forces, so in general there will be four moments. By choosing the point of application of one of them as axis, you can make that moment disappear, but you are still left with the other three.