Normal force at the base of a ladder

[iceice655]

Hello, I was recently tested on finding the normal force at the base of a ladder leaning against a wall as well as its friction force. So this is the question from memory.

Given:
θ the acute angle between the ground and the ladder
μ as the coefficient of friction between the ground and the ladder
M as the mass of the ladder at the center of the ladder, evenly distributed so the inertia if rotated around the center is I=1/12(mL^2), and I = 1/3(mL^2) at the edges
L as the length of the ladder

Known: there is no friction between the wall and the ladder

What is the normal force at the base of the ladder and what is its friction force in terms of θ, μ, g (the gravity constant)M, and L.

 

[erobz]

Hello, I was recently tested on finding the normal force at the base of a ladder leaning against a wall as well as its friction force. So this is the question from memory.

Given:
θ the acute angle between the ground and the ladder
μ as the coefficient of friction between the ground and the ladder
M as the mass of the ladder at the center of the ladder, evenly distributed so the inertia if rotated around the center is I=1/12(mL^2), and I = 1/3(mL^2) at the edges
L as the length of the ladder

Known: there is no friction between the wall and the ladder

What is the normal force at the base of the ladder and what is its friction force in terms of θ, μ, g (the gravity constant)M, and L.

Share your analysis attempt, so it can be examined. For math formatting see the latex guide ( bottom left of the reply window )

 

[iceice655]

Ok what I tried doing was make $\sum{\tau} = 0$ and $\sum{F} = 0$.

Taking the base as the pivot point, I found the perpendicular (to the ladder) force of M at the center to be $mg \cos(\theta)$, and the normal force of the wall on the ladder at the top to be $\mu F_N \sin(\theta)$

Using $\sum{\tau} = 0$, the normal force of the wall on the ladder is at L from the pivot, and the weight of the ladder is at $\frac L 2$, then $\mu F_N \sin(\theta) = \frac 1 2 \cdot Mg \cos(\theta)$, is this correct?

Solving for $F_N$, I would have $F_N = \frac {mg \cos{\theta}} {2 \mu \sin{\theta}}$

 

[PeroK]

Ok what I tried doing was make $\sum{\tau} = 0$ and $\sum{F} = 0$.

Taking the base as the pivot point, I found the perpendicular (to the ladder) force of M at the center to be $mg \cos(\theta)$, and the normal force of the wall on the ladder at the top to be $\mu F_N \sin(\theta)$

Using $\sum{\tau} = 0$, the normal force of the wall on the ladder is at L from the pivot, and the weight of the ladder is at $\frac L 2$, then $\mu F_N \sin(\theta) = \frac 1 2 \cdot Mg \cos(\theta)$, is this correct?

Solving for $F_N$, I would have $F_N = \frac {mg \cos{\theta}} {2 \mu \sin{\theta}}$

The forces involved are largely independent of $\mu$, which is only relevant if you are looking for the maximum distance of the foot of the ladder from the wall.

 

[erobz]

Ok what I tried doing was make $\sum{\tau} = 0$ and $\sum{F} = 0$.

Taking the base as the pivot point, I found the perpendicular (to the ladder) force of M at the center to be $mg \cos(\theta)$, and the normal force of the wall on the ladder at the top to be $\mu F_N \sin(\theta)$

Using $\sum{\tau} = 0$, the normal force of the wall on the ladder is at L from the pivot, and the weight of the ladder is at $\frac L 2$, then $\mu F_N \sin(\theta) = \frac 1 2 \cdot Mg \cos(\theta)$, is this correct?

Solving for $F_N$, I would have $F_N = \frac {mg \cos{\theta}} {2 \mu \sin{\theta}}$

Why does the normal force of the frictionless wall on the ladder involve $\mu$ in your expression? How about a sharing FBD of the ladder as you see it?

 

[PeroK]

Why does the normal force of the frictionless wall on the ladder involve $\mu$ in your expression?

There are only two horizontal forces, so the normal force at the wall must equal the friction force at the ground.

 

[iceice655]

Why does the normal force of the frictionless wall on the ladder involve $\mu$ in your expression? How about a sharing FBD of the ladder as you see it?

So I was thinking in the y-direction, the normal force of the wall on the ladder would equal the friction force at the base of the ladder. (Sorry, touchpads aren't the best writing tools, if there needs to be more clarification in the drawing, I will modify it)

 

[PeroK]

So I was thinking in the y-direction, the normal force of the wall on the ladder would equal the friction force at the base of the ladder. (Sorry, touchpads aren't the best writing tools, if there needs to be more clarification in the drawing, I will modify it)

That diagram looks okay, except why you think the friction force is $\mu Mg$?

 

[iceice655]

That diagram looks okay, except why you think the friction force is $\mu Mg$?

I assumed the friction force at the base was $\mu \cdot F_N$. $F_N$ being that of the ground on the ladder and what I am looking for. Is this correct?

 

[PeroK]

I assumed the friction force at the base was $\mu \cdot F_N$. $F_N$ being that of the ground on the ladder. Is this correct?

No. Friction is a reactive force. It is whatever is needed to maintain equilibrium, up to a maximum possible.

 

[erobz]

There are only two horizontal forces, so the normal force at the wall must equal the friction force at the ground.

Oh, they were jumping ahead. I don’t jump ahead at 6 am.

 

[iceice655]

So at this point I know $F_N$ of the wall on the ladder is equal to the $F_f$ at the base of the ladder. So my equation is $F_f \sin{\theta} = \frac 1 2 Mg \cos{\theta}$. Is it right at this step? My next step should be replacing $F_f$ with the given values, but I am stuck.

 

[PeroK]

You have balanced vertical and horizontal forces. There was no need to look at forces perpendicular to the ladder.

In your OP you mentioned looking at the torque on the system. So, pick a suitable point and calculate the torque about that point.

 

[iceice655]

I chose the point where the ladder and floor touch as my pivot point, I thought this is what I have been doing?
Don't only forces perpendicular to the lever arm result in torque on the ladder?

On another note, looking at just the vertical forces on the FBD, why couldn't I say the $F_N$ of the ground on the ladder is equal to $Mg$?

 

[PeroK]

I chose the point where the ladder and floor touch as my pivot point, I thought this is what I have been doing?

Perhaps I missed that.

Don't only forces perpendicular to the lever arm result in torque on the ladder?

Okay, I see why you wanted those force components. Ultimately, it's the dot product of the force and the vector from the pivot point to the point where the force is applied. The simplest way to calculate this is the force times the perpendicular distance to the force.

On another note, looking at just the vertical forces on the FBD, why couldn't I say the $F_N$ of the ground on the ladder is equal to $Mg$?

I thought you had already done that.

 

[iceice655]

So currently what is incorrect, and what is correct?

Based on what you said, am I right to say
$F_{N (wall on ladder)} = F_{f (ground on ladder)}$ so that
$L \cdot F_{N (wall on ladder)} \sin{\theta} = \frac L 2 Mg \cos{\theta}$

then I need to figure out how to replace $F_{N (wall on ladder)}$ with the given values? This is where I am stuck.

 

[erobz]

So currently what is incorrect, and what is correct?

Based on what you said, am I right to say
$F_{N (wall on ladder)} = F_{f (ground on ladder)}$ so that
$L \cdot F_{N (wall on ladder)} \sin{\theta} = \frac L 2 Mg \cos{\theta}$

then I need to figure out how to replace $F_{N (wall on ladder)}$ with the given values? This is where I am stuck.

$N_f = Mg$ as you suspected as a consequence of the sum of the forces in the vertical direction.

That implies $f_r= \mu Mg$ as you stated

And by sum of forces in horizontal $f_r = N_w \implies N_w \leq \mu M g$

Now apply sum of the torques about the floor contact point( I suppose this should be an inequality to call attention to @PeroK, and @haruspex point about static friction ), and sub for $N_w$ to get $\mu$

Once you have $\mu$ you can get $f_r$

You should be all good to go.

Edit: Geeze, I forgot what the question was! You can get $N_f$ right away.

 

[haruspex]

$N_f = Mg$ as you suspected as a consequence of the sum of the forces in the vertical direction.

That implies $f_r= \mu Mg$ as you stated

Unless I have missed something, and assuming the scenario is static, it only implies $f_r<= \mu Mg$, as pointed out by @PeroK in post #10.
The friction coefficient is provided in order to test the student's understanding of that point.

 

[erobz]

Unless I have missed something, and assuming the scenario is static, it only implies $f_r<= \mu Mg$, as pointed out by @PeroK in post #10.
The friction coefficient is provided in order to test the student's understanding of that point.

Ahh. I see what you’re saying. I’m wrong. Thanks.

 

[erobz]

I thought I saw what you are saying, but now I'm unsure. I thought about PM you and Pero , but given the confusion I'm probably causing the OP it might be best to air this publicly and cut my losses.

I believe it is true that $\mu$ fixes the minimum angle $\theta$ the ladder can be placed in without slipping. So if $\theta \to \pi/2$, $\mu$ is able to go to zero, and $\theta \to 0$, $\mu$ must go to $\infty$.

I get that for any $\theta$, the coefficient of static friction (or "stiction" as @hutchphd prefers):

$$\mu \geq \frac{1}{2}\cot \theta $$

However, given this $\mu$ condition is (at least) minimally satisfied for some angle $\theta$, it must be the case that friction force at the base of the ladder ( when equated to the normal force at the wall) is such that $\sum \tau = 0$ is satisfied, which says to me that:

$$ \frac{l}{2}Mg \cos \theta = l F_{fr} \sin \theta $$

Meaning:

$$ F_{fr} \equiv \frac{1}{2}Mg\cot \theta $$

no more, no less?

 

[haruspex]

$F_{N (wall on ladder)} = F_{f (ground on ladder)}$ so that
$L \cdot F_{N (wall on ladder)} \sin{\theta} = \frac L 2 Mg \cos{\theta}$

then I need to figure out how to replace $F_{N (wall on ladder)}$ with the given values? This is where I am stuck.

I don't understand your difficulty. You have found $F_{N (ground on ladder)}$ and that
$LF_{f (ground on ladder)}\sin{\theta} = \frac L 2 Mg \cos{\theta}$.
Isn’t it trivial to find $F_{f (ground on ladder)}$ from there?