Math

The answer is 4ms^-2

Hi Princess Sophie,
I can't answer your question i am sorry.
There are 2 people here who might be able to. One is 'I like Serena' and the other is DavidQD.
I have left them an attn post to notify them of your question. I am hoping that one of them will answer it tomorrow.
It may have been answered already if you had given it a more descriptive title.
They were both on today, I think that since you only called it 'math' neither of them may have notice it.
Anyway I hope that it is answered tomorrow.

Princess Sophie E:

A uniform ladder of weight w rests on rough horizontal ground against a smooth vertical wall. The vertical plane containing the ladder is inclined at an angle x to the vertical. Prove that if the ladder is on the point of slipping and U is the coefficient of friction between it and the ground, then tanx=wU.

Hi Princess Sophie!

Where the ladder touches the ground it is pushed up by a normal force that is equal and opposite to the force of gravity w.
The friction with the ground is U times this normal force, so that is Uw.
These 2 forces form a right triangle that is the same as the right triangle that the ladder forms with the wall and the ground.
Since the tangent of x is the opposite side divided by the neighboring side, we have that tanx = (Uw) / w = U.
It would appear that what you are supposed to prove (tanx=wU) is not correct. It should be tanx=U.

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Excerpt from Serena's reply:
. These 2 forces form a right triangle that is the same as the right triangle that the ladder forms with the wall
. and the ground.
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The 2 forces that form a right triangle are not the same, they are "mirror" images. The triangles are combined to make a rectangle and it's used to calculate vectored torque forces.
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. Since the tangent of x is the opposite side divided by the neighboring side, we have that
. tanx = (Uw) / w = U.It would appear that what you are supposed to prove (tanx=wU) is not correct. It should be tanx=U.
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It's not that (tanx=wU) is incorrect, it is simply neutralized at the point of equilibrium: w=1 (w/w as you wrote). It continues to apply after it starts falling. It also applies before it reaches equilibrium. If this were not the case then the Kinetic coefficient would not apply. The Static friction (and Kinetic friction) are products of the weight and the friction Coefficient. This is demonstrated by the base formula condensed for the "special" case of equilibrium. This case is noted in the text and formula below.
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Princess Sophie E wrote:A uniform ladder of weight w rests on rough horizontal ground against a smooth vertical wall. The vertical plane containing the ladder is inclined at an angle x to the vertical. Prove that if the ladder is on the point of slipping and U is the coefficient of friction between it and the ground, then tanx=wU.
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A preface of terms and variables for proof. (Some may not be needed here, but they will help as you advance in the subject.)
The wall is perpendicular to the ground.
The weight of the ladder (w) is measured in Newtons (It is found by multiplying its mass in kilograms by the acceleration of gravity (9.8M/s^2) it symbol is N.
There are two types of friction coefficients Static and Kinetic. Static is used here as it the force in effect to the point of slipping. Static friction is the force between two objects that are not moving relative to each other.

Fs=U*N where
Fs = Static friction --> This is the force required to overcome the friction.
U = Static Friction Coefficient, derived by formula or obtained by empirical data.
N = Normal Force in Newtons.
Solve for U: U=Fs/N
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The ladder is at equilibrium when the force of the wall equals force of the ground WaF=GrF:
In this case the ground force is determined by the weight of the ladder times a vector. This vector is a ratio of the downward force and the lateral force at a right angle. The ratio is a tangent function of the ladder's angle (a).
As the ladder's angle increases the lateral force increases by the Tangent (ratio) of its angle.

The formula is Fs*L=w(L/2)Tan(a):The left side of the equation is the force holding the ladder in place. The right side is the force applied to this point. L is divided in half as half the force is the against the wall. The friction of the smooth wall is negligible and not considered here.

Some redefined variables:
Fs = Static friction. It's derived by the product of u times N: where N is the normal force in Newtons (Newtons is a unit of weight. So Fs= u*w)
u is the Static Friction Coefficient
w is the weight (usually defined as N) of the ladder
L is the Length of the ladder

So to include the friction coefficient, rewrite the formula as w*u*L=w(L/2)Tan(a): This is at equilibrium -- the point at which the latter begins to fall. L and w are on both sides of the equation. Factoring these out yields u=(1/2)Tan(a): At this point the distributed "Weight" is equal known as equilibrium.

Different friction Coefficients and/or weights (where the top half or bottom half of the ladder is heaver) will change the angle at which the equilibrium point is reached. That equation w ₁*u*L=w ₂(L/2)Tan(a) implies w ₁d ₁=w ₂d ₂ (weight distance)

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Here are related questions (one with an answer) on the physics described above. These, too, will help you advance in the subject.

A ladder with a mass of 30 Kg and length of 3.5 Meters with rubber feet rests on a dry concrete floor. The static friction coefficient for Rubber/concrete is 0.7. What is the maximum angle the ladder can lean a smooth wall before it slides?

a=arctan(2f). arctan(2 * 0.7)= 54.5 Deg (Note this angle is measured at the top of the ladder, not the bottom -- subtract this from 90 to find the angle relative the floor).

What is the maximum angle if the concrete is wet (friction coefficient drops to 0.55)?

What is the friction coefficient if the ladder is at equilibrium when the angle is 33 degrees?
(This condition might occur if the floor were covered with straw and the rubber feet of the ladder were worn off.)
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