Physics ladder problem help!!! PLEASE!?

Question

An 11.0 m, 230 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground is 0.70, and the ladder makes a 50.0° angle with the ground. How far up the ladder can an 870 N person climb before the ladder begins to slip?
____m

Answer 1

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)

Answer 2

Friction Force,Mu =0.7
Ladder Length, L =11 m
Ladder Set With Respect To Ground At
An Angle =50°
Weight Of Ladder =230 N
Weight Of Man =870 N
Force Exerted By Forces At Base
Of Ladder (Sum Of y's), P (230 + 780) =1100 N
Distance From BaseTo Top (11*Sin(50)) =8.426488874 m
Distance Along Base (11*Cos(50)) =7.070663707 m
Distance From Wall To CG Of Ladder =3.535331853 m
Distance From Wall To CG Of Man =x m
Distance From Base To CG Of Man =y m
Horizontal Force Holding Ladder, H (Mu*P) = 770 N
Take Moments About Point Where Ladder
Touches Wall (Counterclockwise = +)
7.0707*P - 8.4265*H - 3.5353*230 - x*780 = 0
Substituting For H & P
7.0707*1100 - 8.4265*770 - 3.5353*230 - x*780 = 0
Solving For x
x = (7.0707*1100 - 8.4265*770 - 3.5353*230) /780
x =0.547409195m
Distance From Base Of Ladder To CG Of Man,
(7.0707 - x), (7.07071 - 0.547) =6.523254511 m
Distance From Base To CG Of
Man, y = x*Tan(50) = 7.774111999m

Answer 3

Oh, it extremely is a robust one! the single ingredient you already be attentive to is that, for an merchandise to maintain a similar orientation, the sum of the torques from all forces would desire to be equivalent to 0.  what isn't so obtrusive is a thank you to establish this difficulty so which you're watching torques. Neither end of the ladder can transmit any torques without delay.  The wall end can in common terms transmit common forces (horizontal), whilst the floor end can transmit the two common (vertical) and frictional (horizontal) forces.  The frictional tension and the traditional tension from the wall end would desire to be equivalent and opposite by using definition.  The question is, what are the torques from the different forces, and what's a solid place to degree them from? enable's degree from the floor end of the ladder.  it is (6.0m * sin 22° = 2.25 m) from the wall.  the burden of the ladder is 17.0 kg * g, and is a million.a hundred twenty five m from the factor of measurement; the torque is subsequently 19.a hundred twenty five g N-m.  the guy on the ladder is a million.688 m from the factor of measurement, and has a weight of seventy six g N; the torque is subsequently 128.25 g N-m.  The sum of the torques is 147.375 g N-m. This torque would desire to be resisted by using the traditional tension from the wall.  the factor of touch with the wall is (6.0 m * cos 22°=5.56m) from the factor of reference, so the traditional tension would desire to be 147.375 g N-m / 5.fifty six m = 25.8 g N. because of the fact the touch with the wall is frictionless, each and all the burden is exerted on the touch with the floor.  This sums to ninety 3 kg * g.  The minimum coefficient of friction for the ladder to stay placed is: 25.8 g N / ninety 3 g N = 0.277.

Answer 4

Since I don't even start to understand the question, I must ask if you do. Once you understand the question, you should see what formulae are applicable to the situation and all you need to do is plug in the values.

Good luck,

Answer 5

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