Uniform plank resting on two supports.

• annanause
In summary, the problem involves a uniform plank with a weight of 30N and length of 3m, resting on two supports at distances of 0.5m and 2.0m from the left-hand end of the plank. A weight of 18N is suspended from the left-hand end. The reactions at the supports, X and Y, were calculated to be 107N and 196N respectively. To find the amount of weight needed to make the reaction Y zero, the equations of force and moment balance must be used. Initially, only the plank's weight was considered, but it was later added to the moments equation. The final calculations showed that X must be increased by 39N and Y decreased by
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.

Last edited:
Hi an,

annanause said:
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.
annanause said:
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.

BvU said:
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.

annanause said:
mg = 30N x 9.8m/s
The plank has a weight of 30N, not a mass of 30kg.

haruspex said:
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 said:
So I don't need to multiply with 9.8? I can just keep it 30N ?
So mg = 30N ?

annanause said:
So I don't need to multiply with 9.8? I can just keep it 30N ?
Yes.

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

haruspex said:
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

annanause said:
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:
annanause said:
X x 1 = Y x 0.5

haruspex said:
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

annanause said:
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?

haruspex said:
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.

annanause said:
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.

haruspex said:
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?

haruspex said:
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...

annanause said:
X - 30N x 1m = 30Nm
How do you get that? I thought you were taking moments about the mid point of the plank.

haruspex said:
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

annanause said:
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.

haruspex said:
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 :(

annanause said:
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?

haruspex said:
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!

haruspex said:
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

annanause said:
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.

haruspex said:
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 :)

annanause said:
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.

BvU

1. What is the purpose of a uniform plank resting on two supports?

The purpose of a uniform plank resting on two supports is to distribute a load evenly across the length of the plank, allowing it to support heavier objects without breaking or bending.

2. What materials are commonly used for a uniform plank?

Common materials used for a uniform plank include wood, metal, and synthetic materials such as plastic or composite materials.

3. How do you determine the maximum weight a uniform plank can support?

The maximum weight a uniform plank can support is determined by factors such as the type of material used, the length and thickness of the plank, and the distance between the supports. This can be calculated using various engineering principles and formulas.

4. Can a uniform plank resting on two supports be used for outdoor applications?

Yes, a uniform plank can be used for outdoor applications as long as it is made from a material that is able to withstand the outdoor elements, such as pressure-treated wood or weather-resistant metal.

5. Are there any safety precautions to consider when using a uniform plank resting on two supports?

Yes, safety precautions should always be taken when using a uniform plank resting on two supports. It is important to ensure that the plank is properly secured and that the weight being placed on it does not exceed its maximum weight capacity. It is also important to use caution when walking or standing on the plank to avoid slipping or falling.

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