I Solving a Physics Problem: Ladder Facing a Wall

[jasonpeng]

Hey everybody. I just took a test where a problem described, "a ladder is leaning against a wall with an angle of A. It has a length L and it weighs mg. Assume no friction against the vertical wall and a frictional coefficient of B, find Ffriction"

My dad, a quantum physicist, explained how the problem was done, saying that the normal force was mg because the forces in the Y direction added up to 0, the coefficient wasn't needed, and he used Torque (clockwise) = Torque (counterclockwise) to solve for Ffriction in a systems of equations. However, first of all, it wouldn't make sense that there was extra info in the problem, and second of all, I don't understand how the Fn could be mg if it was applied to the end (it just makes logical sense that if it is applied further from the weight center, it affects the translation force less.) I think he may be out of shape in terms of classical physics and forgot the exact way to do this problem and thus I'd like somebody to explain how to ACTUALLY do the problem.

 

[Nidum]

What is the friction force that is needed to act on the foot of the ladder to ensure that the ladder doesn't fall down ?

 

[Chestermiller]

It sounds to me like your dad's analysis is right-on. Have you drawn a free body diagram of the ladder?

 

[jasonpeng]

It sounds to me like your dad's analysis is right-on. Have you drawn a free body diagram of the ladder?

yeah. the normal force is at the bottom of the ladder while the MG is at the middle (center of gravity) so how do those 2 cancel out? if you put a pencil on a table, pushing it in the center to make it translate if way easier than pushin on the edge because it rotates at the same time, so part of the force becomes rotational acceleration, right?

 

[Chestermiller]

yeah. the normal force is at the bottom of the ladder while the MG is at the middle (center of gravity) so how do those 2 cancel out? if you put a pencil on a table, pushing it in the center to make it translate if way easier than pushin on the edge because it rotates at the same time, so part of the force becomes rotational acceleration, right?

In your problem, there is no rotational acceleration. It is in static equilibrium. Do you believe that, for a rigid body in static equilibrium, you must have a balance of forces and a balance of moment? Yes or no?

 

[jasonpeng]

In your problem, there is no rotational acceleration. It is in static equilibrium. Do you believe that, for a rigid body in static equilibrium, you must have a balance of forces and a balance of moment? Yes or no?

I do. However, here what I've considered. If there were no side wall and the floor had no friction, what would happen is that the object would rotate & accelerate CCW but the center of mass would go downwards at the same time. That means the Fnormal is not as much as mg. Once you add the leaning wall, the F(leaning-wall-normal) will push the ladder sideways, but it still rotates and translates downwards. Only when the friction is added, another sideways force, then does the object reach equilibrium. So if the normal force wasn't enough to counteract the mg when there was no ground friction, why would it suddenly increase if you add a ground friction?

essentially what happens is that if there is no friction on the ground, the normal force causes the ladder to rotate and it also keeps falling. If there is a leaning wall, it does the same thing, except the wall will apply a sideways force on the ladder and make it translate sideways while rotating and falling at the same time (the sideways translation keeps the end of the ladder touching the wall in the same X-coordinate). But once the ground friction is added, the leader stops both rotating and translating.

 

[jasonpeng]

What is the friction force that is needed to act on the foot of the ladder to ensure that the ladder doesn't fall down ?

that's what I'm trying to find out. the Ffriction on the ground. If you mean the leaning wall, it has no friction.

 

[Chestermiller]

I do. However, here what I've considered. If there were no side wall and the floor had no friction, what would happen is that the object would rotate & accelerate CCW but the center of mass would go downwards at the same time. That means the Fnormal is not as much as mg. Once you add the leaning wall, the F(leaning-wall-normal) will push the ladder sideways, but it still rotates and translates downwards. Only when the friction is added, another sideways force, then does the object reach equilibrium. So if the normal force wasn't enough to counteract the mg when there was no ground friction, why would it suddenly increase if you add a ground friction?

It doesn't suddenly increase. You are talking about two different problems.

essentially what happens is that if there is no friction on the ground, the normal force causes the ladder to rotate and it also keeps falling. If there is a leaning wall, it does the same thing, except the wall will apply a sideways force on the ladder and make it translate sideways while rotating and falling at the same time (the sideways translation keeps the end of the ladder touching the wall in the same X-coordinate). But once the ground friction is added, the leader stops both rotating and translating.

This is all correct. So?

My recommendation is that you solve both problems so you can compare the results and get a better understanding of their relationship. It should be pretty interesting.

 

[jasonpeng]

so the normal force is bigger if I add a frictional force sideways?

 

[russ_watters]

However, first of all, it wouldn't make sense that there was extra info in the problem...

It is, and it is critical that you get used to this. FrI'm now on, in school and in real life, you may be given extra (or not enough!) information and have to figure out what you need and what you dont.

 

[Chestermiller]

so the normal force is bigger if I add a frictional force sideways?

We can continue speculating about this forever, of we can get down to business and actually solve both versions of the problem to see how this all plays out. I can help you solve the frictionless sliding version of the problem if you are game to try. Are you?

 

[jasonpeng]

We can continue speculating about this forever, of we can get down to business and actually solve both versions of the problem to see how this all plays out. I can help you solve the frictionless sliding version of the problem if you are game to try. Are you?

Yep. Would love to. Let me get my pencil and paper

 

[jasonpeng]

It is, and it is critical that you get used to this. FrI'm now on, in school and in real life, you may be given extra (or not enough!) information and have to figure out what you need and what you dont.

not in the questions from the book's tests though. I mean it would make sense in real life if there was extra info, but not from that book's problems

 

[jbriggs444]

not in the questions from the book's tests though. I mean it would make sense in real life if there was extra info, but not from that book's problems

It is good design to provide extra information in a problem. It helps train the student for real life. And it works to prevent the pattern matching "what formulas do I have that take a distance, a time and a force as inputs" approach to problem solving.

 

[Chestermiller]

not in the questions from the book's tests though. I mean it would make sense in real life if there was extra info, but not from that book's problems

The first step in the analysis is to quantify the kinematics of the ladder motion. We will use the figure below to address that:

From the geometry of this figure, what are the x and y coordinates of the center of mass of the ladder (in terms of L and $\theta$)?

 

[jasonpeng]

The first step in the analysis is to quantify the kinematics of the ladder motion. We will use the figure below to address that:
View attachment 206788

From the geometry of this figure, what are the x and y coordinates of the center of mass of the ladder (in terms of L and $\theta$)?

Well I've found some example problems of the exact same question online, and I know HOW to do it, but I don't know why. How come the normal force is larger when it is in equilibrium but less when there is no friction on the ground if it's the same ladder, same gravity, etc.?

 

[jasonpeng]

The first step in the analysis is to quantify the kinematics of the ladder motion. We will use the figure below to address that:
View attachment 206788

From the geometry of this figure, what are the x and y coordinates of the center of mass of the ladder (in terms of L and $\theta$)?

but for the sliding object: the center of mass is at .5lsin(theta) and .5lcos(theta).

 

[Chestermiller]

Well I've found some example problems of the exact same question online, and I know HOW to do it, but I don't know why. How come the normal force is larger when it is in equilibrium but less when there is no friction on the ground if it's the same ladder, same gravity, etc.?

Like I said (several times), this will all reveal itself when we actually analyze the problem. Until then, we are just waving our hands.

 

[jasonpeng]

Like I said (several times), this will all reveal itself when we actually analyze the problem. Until then, we are just waving our hands.

alright, let's go on then. Sorry for the delay by the way, I'm in a summer camp so I'm busy a lot of the time

 

[Chestermiller]

but for the sliding object: the center of mass is at .5lsin(theta) and .5lcos(theta).

Excellent. Now, using these results, in terms of L, $\theta$, and $d\theta /dt$, what are the x and y components of the velocity of the center of mass?

Please do me a favor. Please use LaTex to do the equations. There is a LaTex tutorial in the Physics Forums help.

 

[jasonpeng]

Like I said (several times), this will all reveal itself when we actually analyze the problem. Until then, we are just waving our hands.

sorry, what's d?

 

[Chestermiller]

sorry, what's d?

You have had calculus, correct?

 

[jasonpeng]

sorry, what's d?

scratch that, could you lead me through how I would find the acceleration of the center of mass? I'm just confuse over how the center of mass moves translationaly if forces are being applied to the object away from the cetner of mass.

 

[jasonpeng]

You have had calculus, correct?

No, I have not. I'm in 10th grade at the moment. this problem showed up in a book I bought for high school physics so I could learn it on my own time

 

[jasonpeng]

Is there a way that the problem can be solved without calculus?

 

[Chestermiller]

Is there a way that the problem can be solved without calculus?

Unfortunately I don't think that the sliding problem can be adequately analyzed without using calculus. Are you familiar with the concepts of angular acceleration, angular velocity, and moment of inertia? In this sliding problem, the acceleration of the center of mass of the ladder, and the angular acceleration of the ladder are not constant, but are functions of time.

 

[jasonpeng]

Unfortunately I don't think that the sliding problem can be adequately analyzed without using calculus. Are you familiar with the concepts of angular acceleration, angular velocity, and moment of inertia? In this sliding problem, the acceleration of the center of mass of the ladder, and the angular acceleration of the ladder are not constant, but are functions of time.

Yes, I've learned precalculus so I do know my angular things.

 

[Chestermiller]

Yes, I've learned precalculus so I do know my angular things.

OK. Fasten your seatbelt.

The sliding motion of the latter is completely described once we know how the angle $\theta$ varies as a function of time t: $\theta=\theta (t)$. The angular velocity of the ladder at time t, $\omega(t)$, is equal to the rate of change of the angle $\theta$ with respect to time, and is represented symbolically (using calculus) by $$\omega = \frac{d\theta}{dt}$$The angular acceleration of the ladder at time t, $\alpha(t)$, is equal to the rate of change of the angular velocity with respect to time, and is represented symbolically (using calculus) by $$\alpha=\frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}$$
Using calculus, we can also obtain equations for the acceleration of the center of mass of the ladder in terms of the angular velocity and angular acceleration. We thereby obtain, for the horizontal and vertical components of the CofM acceleration, the following:
$$a_x=-\frac{L}{2}(\omega^2 \cos{\theta}+\alpha \sin{\theta})=-\frac{L}{2}\left[\left(\frac{d \theta}{dt}\right)^2 \cos{\theta}+\frac{d^2\theta}{dt^2} \sin{\theta}\right]$$
$$a_y=\frac{L}{2}(-\omega^2 \sin{\theta}+\alpha \cos{\theta})=\frac{L}{2}\left[-\left(\frac{d \theta}{dt}\right)^2 \sin{\theta}+\frac{d^2\theta}{dt^2} \cos{\theta}\right]$$
Notice that none of what we have done so far in any way relates to the original static equilibrium friction problem that you were solving. All of this is completely new and different. This is an indication of how very different the two problems are.

This completes what we wanted to do in terms of analyzing the kinematics of the ladder slippage. Now let's get to the dynamics. Please write down the Newton's 2nd law force balance equations for the sliding ladder in the horizontal and vertical directions, in terms of $N_H$, $N_V$, mg, $ma_x$, and $ma_y$.

(Note that this will not yet complete the dynamic equations, because we will also need a moment balance about the center of mass, involving the angular acceleration and moment of inertia of the ladder).

 

[Chestermiller]

@jasonpeng : With respect, I feel that this analysis has gotten a little too complicated, even for a smart 10th grader like yourself. So I am going to fill in some of the intermediate steps, and then let you complete the solution for the reaction forces.

To simplify things, we will confine attention only to the initial state in which we first release the ladder to start sliding. At this initial point in time, the angular velocity of the ladder is zero, and the accelerations of the center of mass in the horizontal and vertical directions become:
$$a_x=-\frac{L}{2}\alpha \sin{\theta}$$
$$a_y=+\frac{L}{2}\alpha \cos{\theta}$$
where now $\theta$ is the initial angle of the ladder and $\alpha$ is its initial angular acceleration.

The Newton's 2nd law force balances on the ladder in the horizontal and vertical directions, and the moment balance on the ladder are given by:$$ma_x=-m\frac{L}{2}\alpha \sin{\theta}=N_H\tag{1}$$
$$ma_y=+m\frac{L}{2}\alpha \cos{\theta}=N_V-mg\tag{2}$$
$$m\frac{L^2}{12}\alpha=N_H\frac{L}{2}\sin{\theta}-N_V\frac{L}{2}\cos{\theta}\tag{3}$$
where $m\frac{L^2}{12}$ is the moment of inertia of the ladder about its center of mass.

Do these relationships in any way make sense to you?

The next step is your turn. Your assignment is to solve these three linear algebraic equations for the three unknowns $\alpha$, $N_H$, and $N_V$ in terms of m, g, and $\theta$. We will then be able to directly compare the results for these reaction forces with those obtained for the ladder problem with friction.

 

[jasonpeng]

coudl you explain how you got the m(l^2/12)alpha part?

 

[jasonpeng]

coudl you explain how you got the m(l^2/12)alpha part?

and how about the translational acceleration? how do the normal forces act on that? and also, isn't alpha also an unknown variable?

 

[Chestermiller]

coudl you explain how you got the m(l^2/12)alpha part?

For a finite rigid body experiencing both translation and rotation, in addition to satisfying the force balances (net force = ma) in the horizontal and vertical directions, one must also satisfy a balance of moments. The balance of moments is the rotational analog of a force balance. It says that the sum of the moments of the forces about the center of mass of the body is equal to the moment of inertia I (analogous to mass) times the angular acceleration (analogous to translational acceleration). That is (net moment = I $\alpha$). For a rigid rod or a rigid ladder, the moment of inertia I is equal to $mL^2/12$. You can look this up online in a Googled table of moments of inertia for various objects. The derivation of the relationship for the moment of inertia is obtained using calculus.

 

[Chestermiller]

and how about the translational acceleration? how do the normal forces act on that? and also, isn't alpha also an unknown variable?

Eqns. 1 and 2 in post #29 describe the translational acceleration. The sum of the external forces are equal to the mass times the acceleration of the center of mass (for the horizontal and vertical directions). Eqn. 3 is the moment balance describing the angular acceleration.

In my post #29, I said "the three unknowns $\alpha$, $N_H$, and $N_V$" ; so, yes, $\alpha$ is also an unknown variable

 

[jasonpeng]

could you also explain the equations 1 and 2 relating Ax with alpha?

 

[Chestermiller]

could you also explain the equations 1 and 2 relating Ax with alpha?

That follows from the geometry/kinematics of the motion. To derive this, you need to take the coordinates of the center of mass at each time, use these equations and calculus to get the velocity of the center of mass, and then do the same thing again to get the acceleration of the center of mass.

 

[scottdave]

Going back to the original problem (not sliding), since it is not moving or accelerating, the sum of the vertical forces = zero, and sum of horizontal forces = 0. How much vertical force does the wall provide on the ladder? How much vertical force does the ground provide on the ladder?

Your father is correct, in this: the coefficient of static friction does not play a role in determining the amount of force, only if there is enough friction available to provide that force.

 

[PaulK2]

Agreeing with Scott, it seems good to solve the original static problem. Chester's diagram on the first page sums it up (although for the way I think about it, I'd add a horizontal force at the foot of the ladder, exerted by the ground, which is what's asked for in the original question).

 

[scottdave]

In order to find the horizontal components of force: summing the moments about a point is the way to go.

 

[Chestermiller]

Agreeing with Scott, it seems good to solve the original static problem. Chester's diagram on the first page sums it up (although for the way I think about it, I'd add a horizontal force at the foot of the ladder, exerted by the ground, which is what's asked for in the original question).

We are going to solve both problems (with and without friction) and then compare the results. The OP is particularly interested in how things change when the frictional force (or externally imposed horizontal force at the base of the ladder) is removed.

 

[alvino]

Isn't this just a statics problem? Sum of forces and sum of moments must be zero if the ladder is resting on a wall and not moving.

Sum of forces (=zero)can allow the relation between the normal force exerted by the wall to the frictional force exerted by the ground(equal to the product of coefficient of static friction and weight).

Sum of moments (about the point on the ladder touching the ground;equals zero)can then allow you to relate the "wall" normal force to the weight of the ladder.

These two combined can give you an expression that relates the coefficient of friction to a geometric expression.

...I think. This is my first post here so I come before you as humble learner. I just read the problem and didn't think it was asking about a dynamic system (especially if OP is in the 10th grade and without calculus).

 

[Chestermiller]

Isn't this just a statics problem? Sum of forces and sum of moments must be zero if the ladder is resting on a wall and not moving.

Sum of forces (=zero)can allow the relation between the normal force exerted by the wall to the frictional force exerted by the ground(equal to the product of coefficient of static friction and weight).

Sum of moments (about the point on the ladder touching the ground;equals zero)can then allow you to relate the "wall" normal force to the weight of the ladder.

These two combined can give you an expression that relates the coefficient of friction to a geometric expression.

...I think. This is my first post here so I come before you as humble learner. I just read the problem and didn't think it was asking about a dynamic system (especially if OP is in the 10th grade and without calculus).

If you check back through the posts, you will see that the OP became intrigued by the dynamic frictionless problem and it's comparison with the static problem. I have tried to accommodate him as much as I can, given his limited mathematical background. This has not been easy.