Poloygon of forces to demonstrate that the crate is in equilibrium

[joe465]

Homework Statement

Use the Polygon of forces to demonstrate that the 200kg crate is in equilibrium on the slope

Homework Equations

F=ma

The Attempt at a Solution

I have attached a picture so you understand the problem a bit better.

I have worked out that the weight acting directly down is:

F=mg
F=200*9.81
F= 1962N

The Force acting back down the slope is:

Wx=W(weight) sin theta
Wx=1962 sin 25
Wx= 829.18N

What other forces are acting on the crate?

Thanks, Joe

 

[tiny-tim]

Hi Joe!

Use the Polygon of forces to demonstrate that the 200kg crate is in equilibrium on the slope
…
The Force acting back down the slope is:

Wx=W(weight) sin theta

What polygon did you get that from?

You're missing the point of the question …

it is asking you to name the three forces on the crate,

then draw a triangle of them,

then use that triangle to calculate the forces (from the given W).

 

[joe465]

My book says:

'Another method of looking at equilibrium involves using the polygon of forces. This
involves taking all the force vectors that are acting on an object and adding them end to
end. If the polygon is completely closed then the object is in equilibrium'

So i thought i had to work out all the forces acting on the crate?

Thanks, Joe

 

[tiny-tim]

My book says:

'Another method of looking at equilibrium involves using the polygon of forces. This
involves taking all the force vectors that are acting on an object and adding them end to
end. If the polygon is completely closed then the object is in equilibrium'

So i thought i had to work out all the forces acting on the crate?

Thanks, Joe

Is this your ICS book again?

If you were given the magnitude and direction of all the forces, then yes could put them in a polygon (in this case a https://www.physicsforums.com/library.php?do=view_item&itemid=99"), see whether it joins up, and if it does then the body is in equilibrium.

But this hardly ever happens …

usually, you're given some of the forces, you're told that it's in equilibrium, and you have to find the remaining forces (so you draw a polygon with an "unknown" side or angle, and then find that side or angle).

In this case, you're told that it's in equilibrium, and you're expected to draw a triangle (with all angles and one side known), find the lengths of the two "unknown" sides, and so find the normal force and the https://www.physicsforums.com/library.php?do=view_item&itemid=39" force.​

(But this is a bad example, since there's really only one https://www.physicsforums.com/library.php?do=view_item&itemid=73" …

the question obviously expects you to split it into the normal and friction components, but doesn't actually say so. )

 

[joe465]

Where would the triangle sit on this occasion?

 

[tiny-tim]

Sit?

You can draw it anywhere, so long as its three sides are parallel to the three directions (vertical, normal, and slope).

 

[joe465]

Hi tiny tim :)

ive attached a picture of an attempt.

Thanks, Joe

 

[tiny-tim]

Hi Joe!

Yes that's the correct triangle, and the correct results.

(But I'm not totally convinced that you got the result the right way …

did you use components of the force, or did you just use geometry?)

 

[joe465]

If I am being totally honest i don't completely know what I am doing haha , it won't be the last time i say that.

The only thing i can tell you is i went in paint and drew some straight lines roughly to what i thought the triangle would look like and worked out the forces by calculating the lenghts of the triangle as written.

If there's a proper way, please enlighten me:)

 

[tiny-tim]

ok, that's correct then!

…drew some straight lines roughly to what i thought the triangle would look like …

yes, that's right, you drew the sides of the triangle parallel to the directions of the three forces

… and worked out the forces by calculating the lenghts of the triangle as written.

yes, you treated this as a geometry problem, which is correct

 

[joe465]

Since i have to use the polygon of forces to demonstrate its in equilibrium will this mean i should do a scale drawing to prove this? or do you know a better method to answer the question

 

[tiny-tim]

You could use a scale drawing (that certainly would work),

but the normal method is to use trigonometry …

the important thing is that you have changed the original physical problem to a geometry problem

(and geometry includes trigonometry)

 

[joe465]

Thanks ever so much, i have one question left and that's pretty much all the maths done for the course so i can relax little:

The crate at the end of the 15m journey is unhooked from the cable. At this point
it is now at the bottom of a conveyor belt which is used to raise the crate to the
next level of the factory at a constant speed.
When the conveyor belt was first installed safety checks were carried out to
determine the limit of angle for safety. It was found that at an angle of 30° the
crates started to slip down the slope.
Calculate the coefficient of friction between the crates and the belt.

Now the book has the most vague coverage of this topic that i don't even have a clue where to start

This is all with the same image first uploaded:)

I greatly appreciate all your help so far, i rather pay money to you than ICS but unfortunately if i ever want out of my current job and to work on better aircraft then i need to get through this:)

 

[tiny-tim]

ok, you can ignore everything up till …

It was found that at an angle of 30° the
crates started to slip down the slope.
Calculate the coefficient of friction between the crates and the belt.

The way you do a problem like this is to assume that it doesn't slip (in other words, it's in equilibrium).

So at angles of 10° 20° or 30° it's in equilibrium, but at 31° it isn't.

If you drew diagrams for each of 10° 20° and 30°,

they would all have the same vertical side, but the other two sides would vary.

(btw, I should have said earlier that you should always label your triangles, for example ABC, so that you can easily refer to them in the equations)

The rule for static https://www.physicsforums.com/library.php?do=view_item&itemid=39" is that the friction force divided by the normal force is always ≤ µ_s (the coefficient of static friction) …

so how would you translate that physical rule into a geometrical rule for the triangle?

 

[joe465]

Thanks,

≤ µs (whats this mean?)

EDIT. just hunted through my course and:

Friction=pressing force*constant

I guess the pressing force is the force of weight and gravity so 1962N
How can i calculate friction without the constant (coefficient of friction?) and
vice versa?

 

[tiny-tim]

Haven't you done https://www.physicsforums.com/library.php?do=view_item&itemid=39" yet?

µ_s is the symbol for the coefficient of static friction

µ_k is the symbol for the coefficient of kinetic (or dynamic or moving) friction

(and the crate is stationary relative to the belt, so the static coefficient is the relevant one)

 

[joe465]

my course shows neither of those symbols:/ and I've scanned through the friction page but its going in one ear and out the other.

How can you calculate friction without the coeeficient, makes no sense that a formula can't be applied without the other

It says that ideally you need to calculate it by trials and without knowing the materials used specifically you can't use the table. its a horrible question haha

 

[tiny-tim]

Yes, the trial basically consists of putting it on a surface, and tilting the surface until the cart starts to slip.

As I said, you calculate the coefficient as equal to the friction force (along the slope) divided by the normal force (perpendicular to the slope).

 

[joe465]

Thanks,

Im not sure why the friction force is is along the slope, is there a direction of the force?

Anyway:

829.18N/1778.18N

Coefficient of friction = 0.466 (thats at 25%)

At 30%:

Slope:
F= w sin theta
F=1962 sin 30
F= 981N

Normal:
F=w cos theta
F=1962 cos 30
F=1699.14N

Coefficient of friction = 981/1699.14
Coefficient of friction = 0.577 (30%)

 

[tiny-tim]

Hi Joe!

(just got up :zzz: …)

Im not sure why the friction force is is along the slope, is there a direction of the force?

Yes, the https://www.physicsforums.com/library.php?do=view_item&itemid=540" is always the direction opposite to which the object "wants" to move.

So the https://www.physicsforums.com/library.php?do=view_item&itemid=39" is always parallel to the surface between two bodies (or to the tangent plane if the surfaces are curved).

At 30%:

Slope:
F= w sin theta
F=1962 sin 30
F= 981N

Normal:
F=w cos theta
F=1962 cos 30
F=1699.14N

Coefficient of friction = 981/1699.14
Coefficient of friction = 0.577 (30%)

Yes, that's fine,

However you could have done it more quickly …

there was no need to find either of the forces …

all you needed was the formula µ_s = Wsin30°/Wcos30° = tan30° = 0.577 ​

 

[joe465]

Morning:), I am not sure how to sattisfy the question here as 30 degrees it says it starts to slip?

 

[tiny-tim]

Yes, for θ ≤ 30° it doesn't slip, and you've found that the ratio of friction/normal force is always tanθ.

The rule for static friction is that, if there's no slipping, friction/normal force ≤ µ_s.

So µ_s is defined as the largest value of friction/normal force for which there's no slipping …

in this case, µ_s = friction/normal force at 30°, ie tan30°. ​