# I Ladder facing a wall

#### Chestermiller

Mentor
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

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

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

Mentor
@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.

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#### 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

Mentor
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.

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#### Chestermiller

Mentor
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

Mentor
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

Homework Helper
Gold Member
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).

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#### scottdave

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

#### Chestermiller

Mentor
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.

A

#### 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).

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#### Chestermiller

Mentor
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.

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