Ladder Against A Wall. Need Physics Help

PHYSICS QUESTION: ladder against a wall...?

When the ladder is in equilibrium,
The downward force mg = the normal reaction at the bottom + the frictional force at the top
Mg = N1 + μ2 N2. ---------------------------------1
Also
Normal force at the top = frictional force at the bottom
N2 = μ1 N1. ----------------------2
Equation 1 is altered as
Mg = N1 + μ1 μ2 N1
Mg = N1 (1+ μ1 μ2) --------------------3
--------------------------
If L is the length of the ladder and θ is the angle of inclination from the vertical,
Taking moment about the bottom point of the ladder,
Mg*(L/2) sin θ. = N2*L cos θ + μ2 N2* L sin θ.
Canceling L through out
Mg sin θ. = 2 N2 (cos θ + μ2 sin θ)
Dividing by cos θ
Mg tan θ = 2 N2 (1+ μ2 tan θ)
From 3 & 2
N1*(1+ μ2 μ1) tan θ = 2 μ1 N1 (1+ μ2 tan θ)
Canceling N1 through out
(1+ μ2 μ1) tan θ = 2 μ1 (1+ μ2 tan θ)
tan θ {1+ μ2 μ1-2 μ1 μ2} =2 μ1
tan θ {1- μ1 μ2} =2 μ1
tan θ =2 μ1/ {1- μ1 μ2}
tan θ =2 * 0.125/ {1- 0.125*0.153}
θ = 14.3°

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Physics ladder problem help!!! PLEASE!?

Let's write some equations...
1. Write a sum of moments equation about the point the ladder of length L touches the wall
let
f- force of friction
W1- weight of the person
W2 - weight of the ladder
R - distance of person from the wall

-L sin(50) f + RW1 + (L/2)cos(50) W2=0
and
f= u N= u(W1+W2)
and let h- height the person climbed then
where h=R tan(50)
we have

R= [(L/2)cos(50) W2) -L sin(50)u(W1+W2)]/ W1
h=[(L/2)cos(50) W2)/W1-L sin(50)u(W1+W2) )]tan(50)

Ladder against a wall. Need physics Help!?

A uniform stationary ladder of length L = 3.1 m and mass M = 15 kg leans against a smooth vertical wall, while its bottom legs rest on a rough horizontal floor. The coefficient of static friction between floor and ladder is μ = 0.35. The ladder makes an angle θ = 52° with respect to the floor. A painter of weight 8M stands on the ladder a distance d from its base.


L = 3.1 m
M = 15 kg
μ = 0.35
θ = 52°

Part (a) Find the magnitude of the normal force N exerted by the floor on the ladder in newtons.

Part (b) Find an expression for the magnitude of the normal force NW exerted by the wall on the ladder.

Part (c) What is the largest value of dmax for which the ladder does not slip, in meters?

Physics problem, a ladder leaning against a frictionless wall?

We assume that the ladder is symmetric. L = 2.94 m is its length
Let Mg be the weight of the ladder (M is its mass and g is gravity).

The torques exerted at equilibrium about the point of contact with the floor must add up to zero. Therefore, the torque of the vertical weight (applied to the center of the ladder) has the same magnitude as the opposing torque of the horizontal force F applied by the wall at the point of contact (there is no vertical component because the wall is frictionless). Therefore, if A is the angle between the floor and the ladder, we have:

(0.5 L cos A) Mg = (L sin A) F
So, F = 0.5 M g / tan A

At equilibrium, the ladder doesn't move horizontally, so the above F is also the value of the horizontal force exerted (in the other direction) by the floor on the ladder.

Let k = 0.398 be the coefficient of static friction between the ladder and the floor. The vertical force between the floor and the ladder is the entire weight Mg of the ladder (because the wall doesn't exert a vertical force). Therefore, there is no slipping when:

F < k Mg
0.5 Mg / tan A < k Mg
0.5/k < tan A

Which means that:

A > arctan (0.5 / 0.398) = 0.8985 rad = 51.48 degrees

Note that the length and the mass of the ladder are irrelevant.
The above holds if the center of gravity of the ladder is in the middle of it. For asymmetrical ladders (those that are narrower at the top, for example) the above coefficient of 0.5 should be modified.

Physics help?

A ladder is leaning against a vertical wall. There is negligible friction between the ladder and the wall, and the coefficient of static friction between the ladder and the ground is μs = 0.135. The ladder is uniform, with its center of gravity located at its geometric center. At what maximum angle α, relative to the wall, can the ladder lean without slipping?

Your help would be very much appreciated. Thanks! ;)

AP Physics ladder problems?

you need to write three conditions of equilibrium.

the first is for the x direction:

in the x direction the forces are the normal horizontal force of the wall against the ladder and the force of friction; since these are the only x forces and the ladder is not accelerating, you know that:

N(hor)-f=0 or N(hor)=f

In the vertical direction, the wall exerts no vertical force (since it is smooth), so the only vertical forces are the weight of the ladder and the vertical normal force of the floor, so we can write:

W-N(vert)=0 or W=N(vert)

Now, you need the equation for torque. to make life a little simpler, let's choose our torques around the point where the ladder touches the floor. This means we know the torques due to friction and N(vert) are zero around this point since they have zero moment (or lever arm) around this point.

This leaves us with the torque due to the N(hor) and the torque due to the weight of the ladder.

We are told the ladder is uniform, so we can act as if all 100N are at the center of the ladder, so the torque due to the ladder is:

W*L/2*cos(theta) where L/2 cos(theta) is the lever arm of the weight

the torque due to the N(hor) is N(hor)*L*sin(theta)

and since these act in opposite directions, they must be equal so:

WLcos(theta)/2=N(hor)Lsin(theta)

notice the Ls cancel, leaving us with:

tan(theta)=W/2*N(hor)

we know W=100N

N(hor)=friction=N(vert)*coeffoffrict
=W*0.4=40N

or tan(theta)=100/80 => theta =51.3 degrees

whew