**Moderators:** MelissaB, kgudger, Ray Trent, Moderators

11 posts
• Page **1** of **1**

Hi there,

our project is about designing a wooden coat stand that can hold 40 kg, its height is 170 cm and have 4 hanger ,,, each hanger must yhold 10 kg. So what is the calculation that we have to do to determine the measurement in the coat stand.

help me plzzz

our project is about designing a wooden coat stand that can hold 40 kg, its height is 170 cm and have 4 hanger ,,, each hanger must yhold 10 kg. So what is the calculation that we have to do to determine the measurement in the coat stand.

help me plzzz

- shy-flower
**Posts:**5**Joined:**Fri Nov 23, 2007 3:10 am

Hi!

Can you specify exactly what you mean by "measurement"? We might be able to answer your question more accurately once we understand the question fully. Thanks!

Can you specify exactly what you mean by "measurement"? We might be able to answer your question more accurately once we understand the question fully. Thanks!

"There is a single light of science, and to brighten it anywhere is to brighten it everywhere." -Isaac Asimov

- staryl13
- Former Expert
**Posts:**404**Joined:**Tue Sep 18, 2007 3:27 pm**Occupation:**Research Assistant**Project Question:**Neuroregeneration**Project Due Date:**N/A**Project Status:**Not applicable

This sounds like an engineering challenge so approaching it like an engineer, What constitutes a (design) failure?

What happens if you hang 10Kg on two adjacent sides and it tips over? Is that a failure?

Is there any penalty for making it heavier or using a larger base than another team?

Is there any peanalty for using more expensive materials than another team?

Is there any benefit from making the coat stand so that it can accomodate a 2 inch displacement of the top from center without tiping over?

What woodworking and wood joining skills does your team have?

These and many other factors can be considered "cost factors" used to evaluate designs.

What happens if you hang 10Kg on two adjacent sides and it tips over? Is that a failure?

Is there any penalty for making it heavier or using a larger base than another team?

Is there any peanalty for using more expensive materials than another team?

Is there any benefit from making the coat stand so that it can accomodate a 2 inch displacement of the top from center without tiping over?

What woodworking and wood joining skills does your team have?

These and many other factors can be considered "cost factors" used to evaluate designs.

-Craig

- Craig_Bridge
- Expert
**Posts:**1297**Joined:**Mon Oct 16, 2006 11:47 am

What grade are you in? Are you taking physics? Have you had "vectors" in any math class?

If one assumes that each of the pieces and joints is strong enough not to bend or deform, one can analyze this using "torque" and force vectors. Try a little searching and see how far you get and then post back with any additional questions.

If one assumes that each of the pieces and joints is strong enough not to bend or deform, one can analyze this using "torque" and force vectors. Try a little searching and see how far you get and then post back with any additional questions.

-Craig

- Craig_Bridge
- Expert
**Posts:**1297**Joined:**Mon Oct 16, 2006 11:47 am

Thank you Craig_Bridge for your help ..

I'm in grade 12 ... this is physics project... I'm not studing in US...

yes I have taken the vectors and the forces long ago in physices

this is what I have done... I hope it's right.

F = m.a

F = 10 * 9.8 = 98 N

w = weight of the coat stand.

If we consider that the coat stand is 12 kg then W= 117.6 N

âˆ‘fx=2(f cos Î¸)-2(f cosÎ¸)=0

âˆ‘fy=-4(f sinÎ¸)-w=0

4(98 sin Î¸) =-117.6

Î¸ =17.46

how can we determine "torque" .... I don't know how detrmine it with out knowing the height of the hunger (R) that help us to find the torque.

I know that the base should be larger than the hunger so it will not tip over.

thnx again...

I'm in grade 12 ... this is physics project... I'm not studing in US...

yes I have taken the vectors and the forces long ago in physices

this is what I have done... I hope it's right.

F = m.a

F = 10 * 9.8 = 98 N

w = weight of the coat stand.

If we consider that the coat stand is 12 kg then W= 117.6 N

âˆ‘fx=2(f cos Î¸)-2(f cosÎ¸)=0

âˆ‘fy=-4(f sinÎ¸)-w=0

4(98 sin Î¸) =-117.6

Î¸ =17.46

how can we determine "torque" .... I don't know how detrmine it with out knowing the height of the hunger (R) that help us to find the torque.

I know that the base should be larger than the hunger so it will not tip over.

thnx again...

- shy-flower
**Posts:**5**Joined:**Fri Nov 23, 2007 3:10 am

If you want the coat stand to be stable and not tip over, then you really don't want any acceleration to be a factor.

Do a little research on torque (should be in your physics book). If not then http://en.wikipedia.org/wiki/Torque or other web resources.

If you design the coat stand to be symmetric and perfectly vertical, then the weight of the stand itself will be balanced. The only tipping forces will be what you hang on the coat stand. If you use "hangers" to hold the coats, then their force will be applied to the "hook" or "peg" at something that approximates a point which will result in a downward force. Since the point at which this downward force is applied will not be the center of the coat stand but at some horizontal distance out from the center of the stand, a torque will be applied to the peg or hook which will in turn apply a torque to the vertical coat stand piece. This torque will then be applied to the base which will attempt to tip over the coat stand.

As long as all of the structural members and joints do not bend or deflect enough to matter, this turns into a problem of interacting levers and the problem is to figure out how this translates to a weight balance on the base modeled as a lever.

Do a little research on torque (should be in your physics book). If not then http://en.wikipedia.org/wiki/Torque or other web resources.

If you design the coat stand to be symmetric and perfectly vertical, then the weight of the stand itself will be balanced. The only tipping forces will be what you hang on the coat stand. If you use "hangers" to hold the coats, then their force will be applied to the "hook" or "peg" at something that approximates a point which will result in a downward force. Since the point at which this downward force is applied will not be the center of the coat stand but at some horizontal distance out from the center of the stand, a torque will be applied to the peg or hook which will in turn apply a torque to the vertical coat stand piece. This torque will then be applied to the base which will attempt to tip over the coat stand.

As long as all of the structural members and joints do not bend or deflect enough to matter, this turns into a problem of interacting levers and the problem is to figure out how this translates to a weight balance on the base modeled as a lever.

-Craig

- Craig_Bridge
- Expert
**Posts:**1297**Joined:**Mon Oct 16, 2006 11:47 am

Hi Grig ...

Thank you very much for your explanation...

if you don't mind .... May I know your email coz I wanna send you the design that I have done to see if there is something worng on it...

Thank you very much for your explanation...

if you don't mind .... May I know your email coz I wanna send you the design that I have done to see if there is something worng on it...

- shy-flower
**Posts:**5**Joined:**Fri Nov 23, 2007 3:10 am

The site rules outlaw direct communication. If you have something that you can't post and you really need help, you can send an email scibuddy@sciencebuddies.org and ask that they forward it.

One of the goals of this site is to capture information that will be useful to others as well as the original investigator (you).

One of the goals of this site is to capture information that will be useful to others as well as the original investigator (you).

-Craig

- Craig_Bridge
- Expert
**Posts:**1297**Joined:**Mon Oct 16, 2006 11:47 am

I recieved a forwarded anonomized email today with an updated drawing. It is similar to your Nov 27 drawing with the following changes:

1) The base was changed to something flat that is "2*L" wide with the post in the center instead of angled legs. L = 25.

2) The angled pegs at the top are length "r" (15 cm)

3) There are two coats modeled as 10 Kg each hanging on two opposing pegs

4) The height of the center post was not shown; however, I'm assuming it was 170 cm based on the initial posting of the requirements.

5) Two F1 force vectors pointing straight down from each of the 10 Kg weights.

6) Two F2 force vectors pointing straight up at an undetermined point close to the end of the base.

You have indicated the weight of the coat stand is 5 kg.

You then show some calculations that don't make much sense to me so I'm going to ignore them and describe a problem with some missing information and then show how I might work through some calculations.

Missing information

Where is the center of gravity of the coat rack? You didn't provide this, so I'm going to assume you have 4 pegs on each of four sides and the base has 4 legs precisely under each peg based on needing to hang 4 coats. I'm also going to assume that the pegs, vertical, and leg pieces are made out of the same square cross section and material. Total material is 4x15 (pegs) + 4x25 (legs) + 170 = 230 cm. If I did the math correctly, this means that the material weighs about 21.75 g/cm. From this we can calculate the center of gravity to be 15 cm up from the base. (Half of 230 cm is 115 cm. 100 cm is in the base, so you need 15 cm of the vertical to get to half of the material). This is a reasonably low center of gravity so the rack is fairly stable with no coat load.

This means that there is a 5 Kg force vector pointing straight down from a point 15 cm up the vertical post of the coat stand. This is considerably different from where you have F2.

worst imbalanced loading problem case

If coats are hung symetcially, the the coat stand is balanced. If you want to figure out the worst case imbalance, you need to look into a single coat and a two coat on adjacent pegs.

single coat case

Assume theta is 30 degress, then 15 cos theta = 15 x .866 = 13 cm. For the coat rack to tip over, the coat rack will have to tip so that it moves 12 cm before it is beyond the end of the leg. min tan tip angle = 12/170 so min angle is 4 degrees. The actual tip angle will be greater because there is a counter balancing force 5 Kg downward force from the center of gravity of the coat stand itself. Note: If you used lighter materials this extra margin would be less. To actually calculate what the tip angle is you need to do some torque calculations based on figuring out where the center of gravity moves to because of the weight of the coat.

two adjacent coats

The effective base and peg distance is now the square root of 2 times what it was in the single coat case or about 70 percent of what it was because of where the center of the coat gravity is. Which is a 17.5 base - 9.1 peg = 8.4 cm is the min tip point (again obtained by ignoring the weight of the coat rack itself). This turns out to be 2.8 degrees if I did the math correctly.

This is not a lot of margin if the floor is carpet or isn't level so you really need to do the torque on the center of gravity calculations to determine how much extra margin you have.

single coat torque calculations

10 Kg * 15 cm = 150 Kg-cm on the peg to post joint. By having four pegs joined at the same point, this is going to be a very weak point. If you reduced the length of the pegs in half you would halve the torque on the joint by half. This would also double the min tip angles and lower the center of gravity and reduce the overall weight.

Torque on the vertical is a bit more complicated. The effective vertical length for torque purposes is 7.5 cm less which comes from 15 sin 30. 15 Kg * 15 cm = X Kg * 162.5 cm = 150 Kg-cm. X = .92 Kg.

To calculate how much deflection the vertical member has with this this loading you are going to have to look up some beam deflection equations and bulk modulus wood propertiies and see if it is going to deflect enough to approach the minimum tip angle. Finding the bulk modulus of a specific wood sample is not going to be easy.

The other approach is to run a test the actual piece. Put blocks of wood supporting a clear span of 162.5 cm and put a weight twice the end load in the middle (1.84 Kg) and measure how much deflection you see. Double the midpoint deflection as an estimate of end point deflection and figure out the equivalent angle using arc sin( deflection / 162.5)

Personally, I would do the physical experiment instead of trying to estimate the correct bulk modulus for a wood sample.

The base joints can be an overlapping tenon joint so the stress density can be significantly less than the peg joints.

Hope this explains one engineering approach to the design calculations.

1) The base was changed to something flat that is "2*L" wide with the post in the center instead of angled legs. L = 25.

2) The angled pegs at the top are length "r" (15 cm)

3) There are two coats modeled as 10 Kg each hanging on two opposing pegs

4) The height of the center post was not shown; however, I'm assuming it was 170 cm based on the initial posting of the requirements.

5) Two F1 force vectors pointing straight down from each of the 10 Kg weights.

6) Two F2 force vectors pointing straight up at an undetermined point close to the end of the base.

You have indicated the weight of the coat stand is 5 kg.

You then show some calculations that don't make much sense to me so I'm going to ignore them and describe a problem with some missing information and then show how I might work through some calculations.

Missing information

Where is the center of gravity of the coat rack? You didn't provide this, so I'm going to assume you have 4 pegs on each of four sides and the base has 4 legs precisely under each peg based on needing to hang 4 coats. I'm also going to assume that the pegs, vertical, and leg pieces are made out of the same square cross section and material. Total material is 4x15 (pegs) + 4x25 (legs) + 170 = 230 cm. If I did the math correctly, this means that the material weighs about 21.75 g/cm. From this we can calculate the center of gravity to be 15 cm up from the base. (Half of 230 cm is 115 cm. 100 cm is in the base, so you need 15 cm of the vertical to get to half of the material). This is a reasonably low center of gravity so the rack is fairly stable with no coat load.

This means that there is a 5 Kg force vector pointing straight down from a point 15 cm up the vertical post of the coat stand. This is considerably different from where you have F2.

worst imbalanced loading problem case

If coats are hung symetcially, the the coat stand is balanced. If you want to figure out the worst case imbalance, you need to look into a single coat and a two coat on adjacent pegs.

single coat case

Assume theta is 30 degress, then 15 cos theta = 15 x .866 = 13 cm. For the coat rack to tip over, the coat rack will have to tip so that it moves 12 cm before it is beyond the end of the leg. min tan tip angle = 12/170 so min angle is 4 degrees. The actual tip angle will be greater because there is a counter balancing force 5 Kg downward force from the center of gravity of the coat stand itself. Note: If you used lighter materials this extra margin would be less. To actually calculate what the tip angle is you need to do some torque calculations based on figuring out where the center of gravity moves to because of the weight of the coat.

two adjacent coats

The effective base and peg distance is now the square root of 2 times what it was in the single coat case or about 70 percent of what it was because of where the center of the coat gravity is. Which is a 17.5 base - 9.1 peg = 8.4 cm is the min tip point (again obtained by ignoring the weight of the coat rack itself). This turns out to be 2.8 degrees if I did the math correctly.

This is not a lot of margin if the floor is carpet or isn't level so you really need to do the torque on the center of gravity calculations to determine how much extra margin you have.

single coat torque calculations

10 Kg * 15 cm = 150 Kg-cm on the peg to post joint. By having four pegs joined at the same point, this is going to be a very weak point. If you reduced the length of the pegs in half you would halve the torque on the joint by half. This would also double the min tip angles and lower the center of gravity and reduce the overall weight.

Torque on the vertical is a bit more complicated. The effective vertical length for torque purposes is 7.5 cm less which comes from 15 sin 30. 15 Kg * 15 cm = X Kg * 162.5 cm = 150 Kg-cm. X = .92 Kg.

To calculate how much deflection the vertical member has with this this loading you are going to have to look up some beam deflection equations and bulk modulus wood propertiies and see if it is going to deflect enough to approach the minimum tip angle. Finding the bulk modulus of a specific wood sample is not going to be easy.

The other approach is to run a test the actual piece. Put blocks of wood supporting a clear span of 162.5 cm and put a weight twice the end load in the middle (1.84 Kg) and measure how much deflection you see. Double the midpoint deflection as an estimate of end point deflection and figure out the equivalent angle using arc sin( deflection / 162.5)

Personally, I would do the physical experiment instead of trying to estimate the correct bulk modulus for a wood sample.

The base joints can be an overlapping tenon joint so the stress density can be significantly less than the peg joints.

Hope this explains one engineering approach to the design calculations.

-Craig

- Craig_Bridge
- Expert
**Posts:**1297**Joined:**Mon Oct 16, 2006 11:47 am

11 posts
• Page **1** of **1**

Return to Grades 9-12: Physical Science

Users browsing this forum: No registered users and 6 guests