Question

An 8.0 m, 200 N uniform ladder rests against a smooth wall.The coefficient of static friction between the ladder and theground is 0.60, and the ladder

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asked 2021-01-28

An 8.0 m, 200 N uniform ladder rests against a smooth wall.The coefficient of static friction between the ladder and theground is 0.60, and the ladder makes a \(50.0^{\circ}\) angle with the ground. Howfar up the ladder can an 800 N personclimb before the ladder begins to slip?
i keep getting 2.85 m but on this site, the answer is 6.15.
am i doing something wrong with the calculations?
can anyone please elaborate this to me? I'd greatly appreciate it.

Answers (1)

2021-01-29

This is a little tricky. You have to write three equations.And you will have three unknowns. The three equations are sumof horiz forces, sum of vert forces, sum of torques. The key is toplace the person a distance "x" up the ladder, assume the staticfriction force is at its limit, and solve.
What forces act on the ladder:
mg down, the weight of theladder
P down, the weight of theperson
W to the left, the normal force fromthe wall
F up, the normal force from thefloor
uF to the right, the static frictionforce from the floor (equal to the coeff times the normal)
The sum of horiz and vert forces then is:
horiz: \(uF = W\)
vert: \(mg + P = F\)
And the torques? Choose a point of rotation. You canchoose any point since the ladder is not moving. I would choose thepoint where the ladder touches the floor.
Now each force provides torque equal to rFsin? where F is the force, r is the distancefrom the point of rotation to the point where the force is applies,and ? is the angle between r and F. In our case, r is zero foruF and F because they are applied at the chosen point of rotation.For the other forces:
mg r is \(\frac{L}{2} (4 m)\) because the force of gravity acts atthe center of the ladder
P r isx, because this is how far the person is from the bottom of theladder
W r is L (8m) because this force acts at the opposite end of theladder
because the ladder is 50 degrees fromthe horizontal, the angle "r" makes with the horizontal forces is 50 deg, while it makes angle 40 deg with the vert forces. So thesum of the torques:
\(mg (\frac{L}{2})\sin40 + Px \sin40 = WL \sin50\)
So here's your three equations:
\(mg(\frac{L}{2}) \sin40 + Px \sin40 = WL \sin50\)
\(uF = W\)
\(mg + P = F\) andyour three unknowns are x, W, F
Solve thelast for F and sub it into the middle one. Sub that one into thefirst one to get rid of W. You end up with:
\(mg (\frac{L}{2}) \sin40 + Px \sin40 = u ( mg + P) L \sin50\)

You can plug in some numbers at this point:
\(\displaystyle{200}\times{\left(\frac{{8}}{{2}}\right)}\times{\sin{{40}}}+{800}\times{x}\times{\sin{{40}}}={0.6}{\left({200}+{800}\right)}\times{8}\times{\sin{{50}}}\)
Crunch the numbers and:
\(514.23 + 514.23 x = 3677\)
\(x = 6.15\) meters

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