What angle should the ladder be set up at to prevent slipping?

[TobyDarkeness]

Homework Statement

A window cleaner uses a ladder of length L and weighs 200N. He places it agaisnt a wall where, at the point of contact the coefficient of static friction, $$\mu$$, is 0.4. Upon the ground is 0.6.

The window cleaner has a mass of 80kg and is 0.8 of the way up the ladder when it begins to slip. At what angle to the ground did he set the ladder up?

Homework Equations

so i know this is a limiting friction question so i need to use F=[text]\mu[/tex]R and the forces should balance finding the normal using some trig i should get the angle.

The Attempt at a Solution

ok so i know the cleaner is 0.8L up the ladder and the weight of the ladder should act downwards 0.5L of the way up, so i have two downward forces of weight in different places on the ladder. I am having trouble putting the free body diagram together. i have a normal force resulting from the contact of the cleaner, two frictional forces one on the wall and one on the ground, (do i need further normal contact forces on the ground and against the wall?). I am not not sure how many total forces i have in order to resolve. could someone please help with the method desperate.

 

[Doc Al]

ok so i know the cleaner is 0.8L up the ladder and the weight of the ladder should act downwards 0.5L of the way up, so i have two downward forces of weight in different places on the ladder.

Good.

I am having trouble putting the free body diagram together. i have a normal force resulting from the contact of the cleaner, two frictional forces one on the wall and one on the ground, (do i need further normal contact forces on the ground and against the wall?).

Where the ladder touches a surface you'll have both a normal force and a friction force.

I am not not sure how many total forces i have in order to resolve.

The contact forces that I just described plus the weights of ladder and cleaner.

Hint: Don't forget to balance torques.

 

[TobyDarkeness]

Thanks allot, should i combine the total weights of the ladder and the cleaner or treat them separable as they are at different locations on the ladder? balance the torques? at the points of contacts when slipping?

 

[Doc Al]

Thanks allot, should i combine the total weights of the ladder and the cleaner or treat them separable as they are at different locations on the ladder?

If you are just balancing forces, location doesn't matter; but location matters when computing torques.

balance the torques? at the points of contacts when slipping?

That would be a good choice.

Hint: Get two force balance equations, for vertical and horizontal forces. Add a torque equation and you'll have all you need.

 

[TobyDarkeness]

Hint: Get two force balance equations, for vertical and horizontal forces. Add a torque equation and you'll have all you need.

thanks again, i have my two force balance equations, but for a torque equation will i not need information about the moment of inertia and or angular accelerations? inertia can be calculated but I am confused how to build a torque equation with the information given.

 

[Doc Al]

thanks again, i have my two force balance equations, but for a torque equation will i not need information about the moment of inertia and or angular accelerations? inertia can be calculated but I am confused how to build a torque equation with the information given.

You won't need to worry about moments of inertia or acceleration--the ladder's in equilibrium. (You're solving for the point just before it starts to slip.) You have all the forces identified (or at least you should); pick an axis and set the net torque equal to zero.

 

[TobyDarkeness]

ah yes that's what i thought initially, i just a little confused when you brought torques up thanks allot this has been really helpful.

 

[TobyDarkeness]

ok so I've resolved horizontally and vertically,

Vertically, 200+80g=Rb+Ra+(80gcos(theta))cos(theta)

Horizontally 0.6Ra=80gcos(theta)sin(theta)+0.4Rb

where Ra and Rb are the contact forces at each end of the ladder ground to wall respectively, the 80gcos(theta) term is the result of the contact force of the climber on the ladder. 0.6 and 0.4 are the coefficients of friction ground to wall respectively.

so i have these two equations (with some simplifying)

200+80g-80g(cos(theta)^2=Rb+Ra
which becomes
200+80g(1-cos(theta)^2)=Rb+Ra
which is

200+80gsin^2(theta)=Rb+Ra vertical resolved
i have simultaneous equations

80gcos(theta)sin(theta)=0.6Ra-0.4Rb
200+80gsin^2(theta)=Rb+Ra

subtracting i get this
200+80g[sin^2(theta)+cos(theta)sin(theta)]=0.4Ra+Rb

not sure where to go next from here or if I've taken a wrong turn.

 

[Doc Al]

ok so I've resolved horizontally and vertically,

Vertically, 200+80g=Rb+Ra+(80gcos(theta))cos(theta)

I don't understand this.

Downward forces: The weight of climber and ladder
Upward forces: Normal force at ground (Ra) and friction force at wall.

Horizontally 0.6Ra=80gcos(theta)sin(theta)+0.4Rb

Horizontal forces: Normal force at wall (Rb); Friction force at ground.

where Ra and Rb are the contact forces at each end of the ladder ground to wall respectively,

I assume Ra and Rb are the normal components of the contact force.

the 80gcos(theta) term is the result of the contact force of the climber on the ladder.

The contact force of the climber on the ladder is just the weight of the climber. Even better: Just treat the 'climber + ladder' as a single composite system.

Redo your force equations. And you still need to add a torque equation into the mix.

 

[TobyDarkeness]

yes i was referring to the normal forces Ra and Rb, i think i got the directions of the frictional and normal forces facing the wrong way in my diagram. treating the climber and ladder as "one" is much easier it appeared i could do so but i was unsure because of the different positions on the ladder. ok this has been really helpful ill get cracking thanks agian.

 

[TobyDarkeness]

wont the normal force from the cleaner be perpendicular to the incline/ladder and therefore won't i need to find its horizontal and vertical components in order to resolve correctly.

 

[TobyDarkeness]

also isn't 80gcos(theta) needed for the normal component of the climber since the weight acts downwards and the normal is perpendicular to the incline.

 

[Doc Al]

wont the normal force from the cleaner be perpendicular to the incline/ladder and therefore won't i need to find its horizontal and vertical components in order to resolve correctly.

Again, I suggest treating the cleaner as being just an 80 kg mass attached to the ladder. No point in treating the cleaner as a separate object. But if you want to treat the cleaner separately, the normal force between cleaner and ladder must exactly equal the cleaner's weight--after all, the ladder is supporting him. (You'll get the same answer either way.)

also isn't 80gcos(theta) needed for the normal component of the climber since the weight acts downwards and the normal is perpendicular to the incline.

If you treat the cleaner/ladder as one object, then the only normal forces you need to worry about are those against the floor and the wall. Much easier that way.

 

[TobyDarkeness]

ok perfect, will that work even though the cleaner isn't positioned at the centre of the ladder?

ok i have these force equations,
(80g+200)[i haven't added them yet as I am using whole number]=Ra+0.4Rb
0.6Ra=Rb
added them i get
(80g+200)=0.4Ra+1.4Rb
for the torques
L(length of ladder)Rbsin(theta)=LRa0.6cos(theta)
which becomes tan(theta)=Ra0.6/Rb

solve simultaneously from here?

 

[Doc Al]

ok perfect, will that work even though the cleaner isn't positioned at the centre of the ladder?

The position of the cleaner only matters when computing torques.

ok i have these force equations,
(80g+200)[i haven't added them yet as I am using whole number]=Ra+0.4Rb
0.6Ra=Rb

Good.

added them i get
(80g+200)=0.4Ra+1.4Rb

Rather than add them, solve them simultaneously for Ra and Rb.

for the torques
L(length of ladder)Rbsin(theta)=LRa0.6cos(theta)

What are you using as your axis? Be sure to include the torque from every force.

 

[TobyDarkeness]

i was using each end of the ladder as an axis for the torques, should i work out the centre of mass instead and use that as my axis and work around there? A torque from every force? ok including the downward weights?

 

[Doc Al]

i was using each end of the ladder as an axis for the torques,

You can't use both ends at the same time! Pick one axis--either end is fine--and write your torque equation.

should i work out the centre of mass instead and use that as my axis and work around there?

Any axis will do. I'd stick with one of the ends.

A torque from every force? ok including the downward weights?

Yes, every force. Including the weights!

 

[TobyDarkeness]

ah i get taking moments around a point, I am with you now.

 

[TobyDarkeness]

like taking moments i should say

 

[TobyDarkeness]

ok i have my Rb and Ra done, for the torques i have

0.5L*200sin(theta)+(80gsin(theta)*0.8L)=Rb*Lsin(theta)+(0.4Rbsin(theta)*L)

but after simplifying this i divide and lose my angle so I am sure something is wrong.

 

[Gokul43201]

You need to take your time and be more descriptive: for instance, when talking about torques, you need to specify which point (axis) you are calculating the moments about.

Hint: What is the angle theta measured between? (re-check your sines and cosines)

 

[TobyDarkeness]

ok sorry, that was a bit vague, the axis is where the ladder meets the ground, and the angle is between the ladder and the ground.

 

[Gokul43201]

Are the forces that contribute a moment (about the bottom end of the ladder) all acting vertically, or do one or more of them act horizontally?

 

[TobyDarkeness]

there is a force vertically and one horizontally, i left these forces out as they are at the axis point, should i not have? if so what do i take as the "R" value in the torque equation?

 

[Gokul43201]

Those two forces indeed contribute no torque.

I was asking you about the remaining 4 forces. Look at them again: what are their directions?

 

[Doc Al]

ok i have my Rb and Ra done, for the torques i have

0.5L*200sin(theta)+(80gsin(theta)*0.8L)=Rb*Lsin(theta)+(0.4Rbsin(theta)*L)

but after simplifying this i divide and lose my angle so I am sure something is wrong.

OK, you did a good job identifying the forces, but (as pointed out by Gokul43201) you messed up the angles. I assume you're trying to use the formula:

$$\tau = \vec{r}\times\vec{F} = rF\sin\theta$$

The angle theta is the angle between the vector r and the vector F. The vector r points from the axis to the point of application of the force. You'll need a good diagram showing each force and the direction in which it points. Then you can figure out the angle between those vectors. Give it another shot.

Another way to think of $r \sin\theta$ is as the perpendicular distance between the axis and the line of action of the force. This is often called the moment arm or lever arm. Thus torque = Force X lever arm. This might be easier to conceptualize.

 

[TobyDarkeness]

Ah i see, it should fall into place now. Thank you both this has been really helpful learning tool.

 

[TobyDarkeness]

think i got it, simplified to a tan(theta) equation got an angle of 47.5. thanks again.