Determine the perpendicular distance for couple

[goldfish9776]

Homework Statement

two 50N are applied to the corner B and D , determine the moment of couple formed by the two forces by resolving each force into the horizontal and vertical component and adding the two resulting couples . Use the result to determine the perpendicular distance between the line BE and DF

Homework Equations

The Attempt at a Solution

i gt only Bx = 50sin50 = 38.3N (right)
By = 50cos50 = 32.1N ( down)

Dx= 50 sin50 = 38.2 (left)
Dy = 50 cos50 = 32.1N (up)

how to get the perpendicular distance ?
btw , i have used another method which is not relevant to the question asked , whoch is tan 50 = O / 300
O= 357.5mm
hence , perpendicular deistance = 500-357.5mm = 142.5mm

 

[SteamKing]

Homework Statement

two 50N are applied to the corner B and D , determine the moment of couple formed by the two forces by resolving each force into the horizontal and vertical component and adding the two resulting couples . Use the result to determine the perpendicular distance between the line BE and DF

Homework Equations

The Attempt at a Solution

i gt only Bx = 50sin50 = 38.3N (right)
By = 50cos50 = 32.1N ( down)

Dx= 50 sin50 = 38.2 (left)
Dy = 50 cos50 = 32.1N (up)

how to get the perpendicular distance ?
btw , i have used another method which is not relevant to the question asked , whoch is tan 50 = O / 300
O= 357.5mm
hence , perpendicular deistance = 500-357.5mm = 142.5mm

This last calculation does not give the perpendicular distance between the lines of the two forces.

The distance O = 357.5 mm is measured along the edge of the plate, and so is the difference 500 - 357.5 = 142.5 mm.

You must find the distance between the line of each force measured perpendicular to each, similar to what is shown below:

 

[goldfish9776]

This last calculation does not give the perpendicular distance between the lines of the two forces.

The distance O = 357.5 mm is measured along the edge of the plate, and so is the difference 500 - 357.5 = 142.5 mm.

You must find the distance between the line of each force measured perpendicular to each, similar to what is shown below:

CAN YOU GIVE SOME HINT HOW TO FIND ?

 

[SteamKing]

CAN YOU GIVE SOME HINT HOW TO FIND ?

Follow the original directions stated in the problem. Don't go straying off topic.

 

[goldfish9776]

Follow the original directions stated in the problem. Don't go straying off topic.

how to find ? i have only
Bx = 50sin50 = 38.3N (right)
By = 50cos50 = 32.1N ( down)

Dx= 50 sin50 = 38.2 (left)
Dy = 50 cos50 = 32.1N (up)

if i sum up them , i would get 0 .

 

[SteamKing]

how to find ? i have only
Bx = 50sin50 = 38.3N (right)
By = 50cos50 = 32.1N ( down)

Dx= 50 sin50 = 38.2 (left)
Dy = 50 cos50 = 32.1N (up)

if i sum up them , i would get 0 .

You're not supposed to sum them up.

Read the problem statement again. Does it tell you to sum these forces up? Or does it ask you to do something else?

 

[goldfish9776]

You're not supposed to sum them up.

Read the problem statement again. Does it tell you to sum these forces up? Or does it ask you to do something else?

well , force the force at B , i have moment = 50sin50(300) -50cos50(500) = -4560N
for the force at D , i also have moment = 50sin50(300) -50cos50(500) = -4560N

so i would get -9120Nmm / 50 = = 182.4mm , am i right ?

 

[theodoros.mihos]

You can calculate the stress by axes and the total couple force stress as $F\times{D}$. Must be equals.

 

[SteamKing]

well , force the force at B , i have moment = 50sin50(300) -50cos50(500) = -4560N
for the force at D , i also have moment = 50sin50(300) -50cos50(500) = -4560N

so i would get -9120Nmm / 50 = = 182.4mm , am i right ?

You pick one point and calculate the moments of the force components about that point.

If you choose either B or D as the point about which to calculate the total moment, your calculations are greatly simplified.

 

[goldfish9776]

You pick one point and calculate the moments of the force components about that point.

If you choose either B or D as the point about which to calculate the total moment, your calculations are greatly simplified.

so the ans = 500(32.1)-300(38.3)=4560Nmm
4560Nmm / 50 = 91.2 mm ?

 

[SteamKing]

so the ans = 500(32.1)-300(38.3)=4560Nmm
4560Nmm / 50 = 91.2 mm ?

Yes, that is correct.

 

[goldfish9776]

Yes, that is correct.

why not 91.2 x 2 = 184 mm ? this is beacuse when we calculate the sum of couple moment , we will need to consider both side ...
so for the diagram below , the sum of couple moment is F1(d/2 ) +F1(d/2 )

so for the question i asked , the 91.2 mm that we gt is only the d/2 only

 

[SteamKing]

why not 91.2 x 2 = 184 mm ? this is beacuse when we calculate the sum of couple moment , we will need to consider both side ...
so for the diagram below , the sum of couple moment is F1(d/2 ) +F1(d/2 )

so for the question i asked , the 91.2 mm that we gt is only the d/2 only

Yes, but if you calculate the moments of each force about the midpoint of the distance separating them, then M = F * d/2 + F * d/2 = F * d, since both forces are of equal magnitude.

If F = 500 N and M = 4560 N-mm, what must the distance d be equal to?

 

[goldfish9776]

Yes, but if you calculate the moments of each force about the midpoint of the distance separating them, then M = F * d/2 + F * d/2 = F * d, since both forces are of equal magnitude.

If F = 500 N and M = 4560 N-mm, what must the distance d be equal to?

the d = 91.2mm , but the d here that we gt is only half of the perpendicular distance of the question i asked originally , right ?

 

[SteamKing]

the d = 91.2mm , but the d here that we gt is only half of the perpendicular distance of the question i asked originally , right ?

No, it's not. You still have two forces creating the couple, as I showed you in Post #13.

Take the small diagram you have in Post #12 and calculate moments about one end of d (it doesn't matter which end) and again about the center of d. Each moment you calculate should have the same magnitude.

 

[goldfish9776]

No, it's not. You still have two forces creating the couple, as I showed you in Post #13.

Take the small diagram you have in Post #12 and calculate moments about one end of d (it doesn't matter which end) and again about the center of d. Each moment you calculate should have the same magnitude.

ok , i think i gt your point