Calculating Reaction Forces on Cylinders Resting on an L Profile

[Epoch]

Homework Statement

Two cylinders, each with a mass of 10kg, are lying on an L profile.
Neglect the thickness and weight of the L profile.
Calculate the reaction force in b and e.

Homework Equations

G = mg

The Attempt at a Solution

So I tried to make a free body diagram of the cylinders.
I assume their mass is combined, since they are close together.
I took 1 cm = 40N
G = 10*2*9.81 = 196.2 N

I think I found Fxb, but how to find the rest?
Can someone help me?

The answers on my answer sheet are:
Fxb = 100 N
Fyb = 136.14 N
Fye = 36.96 N

 

[haruspex]

I think I found Fxb, but how to find the rest?
Can someone help me?

Per forum rules, you need to post your attempt or, at the least, some thoughts on the matter.
In your answer for Fxb you seem to be using g=10m/s², not 9.8 as you had indicated.

 

[Epoch]

Per forum rules, you need to post your attempt or, at the least, some thoughts on the matter.
In your answer for Fxb you seem to be using g=10m/s², not 9.8 as you had indicated.

That 10 is 10kg.
I don't know if you can take the mass together.
G = 10kg * 2 * 9.81m/s^2 = 196.2 N

 

[Epoch]

Per forum rules, you need to post your attempt or, at the least, some thoughts on the matter.
In your answer for Fxb you seem to be using g=10m/s², not 9.8 as you had indicated.

I made a new drawing.
This time is 1 cm = 50N.

I think I found Fxb, not sure, since it came out 100N
1cm = 50N
2cm = 100N
But Fyb is wrong and I don't understand the force on e.

 

[haruspex]

That 10 is 10kg.

No, I'm saying that to get the answer 100N you must be using g=10m/s².

I don't know if you can take the mass together.

You can in this case. There are three unknowns, the magnitudes of the three reaction forces, and three equations available: force balance in the x and y directions, and moments about some axis. If there were three balls then combining their masses would not give you enough equations to find the forces at the points where the balls contact the L.
Pick an axis b or e, and take moments. [this is edited from what I wrote before, where I mistakenly thought you need the reactions at c and d.]

1cm = 50N

Ok, so you are finding the answers by measuring distances in a drawing, right? That explains why your 100N doesn't match your choice of value for g. Are you expected to find the answers that way? It would be better to use algebra.

 

[Epoch]

No, I'm saying that to get the answer 100N you must be using g=10m/s².

You can in this case. There are three unknowns, the magnitudes of the three reaction forces, and three equations available: force balance in the x and y directions, and moments about some axis. If there were three balls then combining their masses would not give you enough equations.
Pick an axis, a point of contact of one of the balls with the L, say, and take moments.

Ok, so you are finding the answers by measuring distances in a drawing, right? That explains why your 100N doesn't match your choice of value for g. Are you expected to find the answers that way? It would be better to use algebra.

If I follow this
G = m * g
G = 20 * 10 = 200N

N = m * g * cos(30)
N = 20 * 10 * cos(30) = 173.2 N

f = m * g * sin(30)
f = 20 * 10 * sin(30) = 100N
So the X (Fxb) component is right, but why is my Y (Fyb) component wrong?

 

[haruspex]

View attachment 210663

If I follow this
G = m * g
G = 20 * 10 = 200N

N = m * g * cos(30)
N = 20 * 10 * cos(30) = 173.2 N

f = m * g * sin(30)
f = 20 * 10 * sin(30) = 100N
So the X (Fxb) component is right, but why is my Y (Fyb) component wrong?

Since you are combining the masses, you must combine all the forces acting on them. That includes the reaction at e. That provides some of the y force, it does not all come from b.

You need another equation. See my (edited) post #5.

 

[Epoch]

Since you are combining the masses, you must combine all the forces acting on them. That includes the reaction at e. That provides some of the y force, it does not all come from b.

You need another equation. See my (edited) post #5.

You mean that the force of:
cylinder 1 + cylinder 2 + force on "e" = total force on Y ?

Fcylinder 1 = Fcylinder 2 = m * g * cos(30) = 10 * 10 * cos(30) = 86.6 N

total force on Y = 173.2 N = both cylinders?
What I do is clearly wrong since the force for both cylinders should be less that 86.6 N, otherwise there is no place for Fye.
Of course if I understood correctly what you meant.

 

[haruspex]

cylinder 1 + cylinder 2 + force on "e" = total force on Y ?

I don't know what you mean by "force on Y".
Since you are not asked for the reaction forces at c and d, you can treat the cylinders plus the L piece as a single body.
It is supported by a force at e in the Y direction and forces in both X and Y directions at b.
What balance of forces equation does that give you in the Y direction?
Choose an axis, b or e, for finding moments.
What equation can you write for balance of torques about that axis?

 

[Epoch]

I don't know what you mean by "force on Y".
Since you are not asked for the reaction forces at c and d, you can treat the cylinders plus the L piece as a single body.
It is supported by a force at e in the Y direction and forces in both X and Y directions at b.
What balance of forces equation does that give you in the Y direction?
Choose an axis, b or e, for finding moments.
What equation can you write for balance of torques about that axis?

What I meant with the total force is that when I:
Fyb = 136.14 N
Fye = 36.96 N

N force = Fyb + Fye = 173.1N

There is a Y force working on b and e.
I don't understand what you mean with taking a point and finding moments.
Do you mean:

(If I treat the L profile as a whole)
-200N * 0.3m + Fye * 0.6m = 0
-60Nm + Fye * 0.6m = 0
Fye = 60Nm / 0.6m = 36m

Fyb = 173.1N - Fye
Fyb = 173.1 - 36 = 137.1N
?
So the normal force is working on the whole L profile, but it consists of the forces on b and e?

 

[haruspex]

N force = Fyb + Fye = 173.1N

That is right, but it seems to be quite different from what you wrote in post #8.

-200N * 0.3m + Fye * 0.6m = 0

That is the sort of thing, but the equation you have written is quite wrong. The mass centre of the cylinders+L system is not 0.3m up the slope. For writing the correct equation it will be simpler to treat the two cylinders separately.

Technically, you should also have taken into account that the weights and the normal force at e are vertical, not orthogonal to the distances along the L. So each term should also have a factor cos(theta). However, since that is the same for all the forces in the equation those factors cancel.

 

[Epoch]

That is right, but it seems to be quite different from what you wrote in post #8.

That is the sort of thing, but the equation you have written is quite wrong. The mass centre of the cylinders+L system is not 0.3m up the slope. For writing the correct equation it will be simpler to treat the two cylinders separately.

Technically, you should also have taken into account that the weights and the normal force at e are vertical, not orthogonal to the distances along the L. So each term should also have a factor cos(theta). However, since that is the same for all the forces in the equation those factors cancel.

Two cylinders separately: (hypotenuse)
-100N * 0.09m - 100N * 0.27m + Fye * 0.6 = 0
-9Nm - 27Nm + Fye * 0.6m = 0
Ge = 36Nm / 0.6m = 60N
So this is the gravitational force on e?
So to get the normal force on e I need to: 60N * cos(30) = 51.96N
What am I doing wrong?

 

[haruspex]

-9Nm - 27Nm + Fye * 0.6m = 0

I was wrong to say you could get away with omitting the sin(theta) terms. I was forgetting that your Fye is not vertical.
So let's do this properly. The 100N weights are vertical, but the distances you have used, 0.09m and 0.27m, are not horizontal. For taking moments, you need to use the distance from the axis to the line of action of the force. E.g. what is the horizontal distance from the axis b to the vertical line through the mass centre of the lower cylinder?

 

[Epoch]

I was wrong to say you could get away with omitting the sin(theta) terms. I was forgetting that your Fye is not vertical.
So let's do this properly. The 100N weights are vertical, but the distances you have used, 0.09m and 0.27m, are not horizontal. For taking moments, you need to use the distance from the axis to the line of action of the force. E.g. what is the horizontal distance from the axis b to the vertical line through the mass centre of the lower cylinder?

-100N * 0.041m - 100N * 0.205m + Fye * 0.5196m = 0
-4.1 - 20.5 + Ge * 0.5196 = 0
Ge = 24.6 / 0.5196 = 47.34N
Fye = 47.34 * cos(30) = 40.99N

I have the horizontal distance, but I still have it wrong.
Is there something wrong with my equation?

 

[SammyS]

-100N * 0.041m - 100N * 0.205m + Fye * 0.5196m = 0
-4.1 - 20.5 + Ge * 0.5196 = 0
Ge = 24.6 / 0.5196 = 47.34N
Fye = 47.34 * cos(30) = 40.99N
View attachment 210890
I have the horizontal distance, but I still have it wrong.
Is there something wrong with my equation?

You're missing the contribution from the x component of F_e .

 

[haruspex]

You're missing the contribution from the x component of F_e .

The X coordinate is up the slope. The support at e is a roller, so no force there in the X direction.
I guess you meant the horizontal component.

 

[haruspex]

Fye * 0.5196m

Why have you changed the distance in this term? The force F_ye is perpendicular to the 0.6m distance.

 

[Epoch]

Why have you changed the distance in this term? The force F_ye is perpendicular to the 0.6m distance.

You mean like this then?
-100N * 0.041m - 100N * 0.205m + Fye * 0.6m = 0
-4.1 - 20.5 + Fye * 0.6 = 0
-24.6 + Fye * 0.6 = 0
Fye = 24.6 / 0.6 = 41

But why do I need to use the adjacent for the two forces on the cylinders and the hypotenuse for "e"?
I used 0.5196m for "e" to get the force perpendicular with the adjacent because I was using it for the both cylinders.

 

[haruspex]

But why do I need to use the adjacent for the two forces on the cylinders and the hypotenuse for "e"?

Because the force from e is in the y direction, so is normal to the 0.6m from the axis. The two weight forces are in the vertical direction.

 

[Epoch]

Why have you changed the distance in this term? The force F_ye is perpendicular to the 0.6m distance.

I think I figured out my mistake, ignore post #18.

Because the force from e is in the y direction, so is normal to the 0.6m from the axis. The two weight forces are in the vertical direction.

I figured out that my horizontal distances were wrong, I forgot that the L profile width should be neglected.
What you said is totally right, I just did it in 2 steps.

The first time I calculated the G force for both cylinders and e and the took the cosine of Ge = 37.2 N.

Your equation is better since I could have spared a lot of time.
-100N * 0.03293m - 100N * 0.19033m + Fye * 0.6 = 0
-22.326 + Fye * 0.6 = 0
Fye = 22.326 / 0.6 = 37.2N
Fyb = Nforce - Fye = 173.2N - 37.2N = 136N

I think I got it right this time, the only thing is the difference in decimals.
Answer sheet: 36.96 N
My answer: 37.2N
Maybe because I rounded my numbers.

Also I calculated my horizontal distances by measuring with a ruler and putting them up to scale for the drawing.
Example:
The first horizontal distance I measured was 0.45 cm.
The hypotenuse is 7.1cm = 600mm on my drawing.
So I divided 7.1 cm by 0.45cm and got 15.77.
Then I divided 600mm by 15.77 to get the hypotenuse of 38.04mm for my first horizontal distance.
Then I did 38.04 * cos(30) = 32.94 mm, to get the first horizontal distance.
The same applies for the second horizontal distance.

This looks very unprofessional and my question is:
Is there a way to do this properly without measuring?

 

[haruspex]

Is there a way to do this properly without measuring?

Yes. What function of an angle in a right angled triangle is the ratio of the adjacent side to the hypotenuse.

 

[Epoch]

Yes. What function of an angle in a right angled triangle is the ratio of the adjacent side to the hypotenuse.

That would be the cosine = A / H.

I need for example the blue triangle.
I know the hypotenuse of the red triangle, since 180mm / 2 = 90 mm.
But the hypotenuse of the blue triangle is for me unknown.

At first I was thinking if I can do this:

Since the cylinders are round, their diameter is from all sides the same and since we neglect the width of the L profile I only need to measure the angle of the red triangle and use tan = Opposite / adjacent.

 

[haruspex]

That would be the cosine = A / H.

I need for example the blue triangle.
I know the hypotenuse of the red triangle, since 180mm / 2 = 90 mm.
But the hypotenuse of the blue triangle is for me unknown.
View attachment 211035

At first I was thinking if I can do this:
View attachment 211036
Since the cylinders are round, their diameter is from all sides the same and since we neglect the width of the L profile I only need to measure the angle of the red triangle and use tan = Opposite / adjacent.

Some more labels...
Call the point on the same horizontal height as b and directly below c point f.
In between b and f, the point below the centre of the lower cylinder is g.
Can you figure out distance fg?