One way to put a ladder in place on a wall is to \"walk\" the ladder to the wall

Question

One way to put a ladder in place on a wall is to "walk" the ladder to the wall. One end of the ladder is held fixed (perhaps at the base of the wall), and a person slides his hands along the ladder as he walks toward the fixed end as sketched in the figure below. Anyone who has raised a ladder in this way knows that the amount of force the person must apply grows larger initially as the person moves toward the end that is fixed. The length of the ladder is 10 m. The mass of the ladder is 70 kg. The height of the person is 2 m. Assuming the force from the person is perpendicular to the ladder, find the force the person must apply to hold the ladder at an angle of 13

Explanation / Answer

Assumptions and variables:

h: height at which force is applied
x: distance along the length of the ladder at which the force is applied
l: total length of the ladder
?: angle the ladder makes with the horizontal (here I'm assuming the angle given in the problem is the angle with the horizontal, as this was not explicitly stated)
m: mass of the ladder
?_f: torque due to the force of the person (read "tau sub f")
?_g: torque due to the force of gravity (read "tau sub g")

f: force exerted by the person (the answer)

With these assumptions, we derive the following useful relation:

x*sin(?) = h (this should be clear from a simple sketch of the situation)
x = h/sin(?)

At this point, we introduce to concept of torque, the rotational analog of force. This is useful because the ladder is fixed at the base and thus can be treated as a rod rotating around a fixed point. Using torque, we can ignore the force of the ground on the ladder at the base because torque is proportional to the distance from the point of rotation, and the base is at a distance of 0 (because it is the point of rotation).

?_f = x*f*sin(?)

where ? is the angle between the force vector with magnitude f and the position vector from the base of the ladder to the point where the force f acts with magnitude f. Torque is proportional to x, the distance along the ladder at which the force f acts.

Since the force of the person is always perpendicular to the ladder, we know that ? = ?/2 and

?_f = x*f*sin(?/2) = x*f

The torque due to gravity is:

?_g = m*g*(l/2)*sin(?)

where ? is the angle between the gravity force vector and the ladder, which is the same as the angle between the ladder and the vertical. Therefore, ? = ?/2 - ? and

?_g = m*g*(l/2)*cos(?)

here the magnitude of the position vector is l/2 because gravity acts on the center of mass of the ladder, at distance l/2 along the length of the ladder.

Finally, since the ladder is stationary at the angles in question, we know that the sum of the forces acting on the ladder is equal to zero. Therefore, since ?_g and ?_f are in opposite directions, we can equate these two torques and solve for f:

f = m*g*(l/2)*cos(?)/x

From above, we know that

x = h/sin(?)

so

f = m*g*(l/2)*cos(?)/(h/sin(?)) = m*g*l*sin(?)*cos(?)/(2*h)

Note: this equation is only valid for ? >= arcsin(h/l), as anything smaller than this would place the ladder "inside" the person, which makes no physical sense.

As one always should when finished with a physics problem, we check a few easy-to-check cases which we know the answer to in order to verify our result:

Case: ladder is vertical: ? = ?/2, f = 0

This makes sense. When the ladder is vertical it is, ideally, balanced, so no force is required to hold it up.

Case: h=0, f = ?

This may seem like it invalidates our expression, but consider what happens as h gets smaller and smaller.

sin(?) = h/x

sin(?) is constant in our discussion of h, so as h decreases, so does x, pushing the hypothetical force closer and closer to the point of rotation. As the point at which the force is applied approaches the point of rotations, the torque required to hold the constant mass of the ladder approaches infinity, thus f approaches infinity as h approaches 0.

Now that we have decent confidence in our result, simply evaluate f = m*g*l*sin(?)*cos(?)/(2*h) at the two values of ? listed in the problem.

I hope this helps, but keep in mind that in order to fully evaluate the force expression it is necessary to know the mass of the ladder, the length of the ladder, and the height at which the force acts (the height of the person, presumably). If these were not stated somewhere in the problem, then this is a bad problem, and it should be brought up with your instructor (assuming this is for a class).

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