Solving a Physics Problem: Ladder Facing a Wall

[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.