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A ladder of length L = 2.7 m and mass m = 24 kg rests on a floor with coefficient of static friction μs = 0.52. Assume the wall is frictionless. A person with mass M = 61 kg now stands at the very top of the ladder.

1. What is the normal force the floor exerts on the ladder? 833.85N
2. What is the minimum angle to keep the ladder from sliding? 43.877 degree; isn't this should be the same no matter how big the mass is on the ladder?



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A meterstick (L = 1 m) has a mass of m = 0.183 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.

4. What is the tension in the string when the meterstick is vertical?
and I wonder if the angular acceleration is constant at all time?

Western, for the first one, it looks like you solved part 2 as though the man was standing in the center of the ladder. In that case, yes, it should be the same answer no matter how heavy the man is.

However, he's not standing in the center. He's standing at the top.

Therefore, when you compute the total clockwise torque on the ladder it will be from the weight of the ladder, plus from the weight of the man at the top (I think this is double what you computed for the torque contribution from the man). The max horizontal force before the ladder slips is what it appears you calculated, N. Therefore, at the minimum angle before slippage, the wall will be imparting an outward force of 433.602N, and a counter-clockwise torque of . Just set the clockwise and counter-clockwise torques equal to each other, and solve for . I get an answer closer to 59 degrees. And the answer would change depending on the man's weight (for example, if he weighed nothing, the correct answer would be the one you got before).

For the second question I have to guess a little because you skipped the first 3 parts (I think). The string on the right must have broken?

Anyway, when the meter stick is vertical the tension in the string will be the weight of the meter stick (0.183 * 9.81), plus the net force due to centripetal acceleration. The 25 cm of the stick that is pointed upwards would result in a small upward force, while the 75 cm of the stick that is pointed downward would result in a much larger (9x as large I think) downward force.

The angular acceleration should NOT be constant, since gravity (pointing downward at all times) is responsible for the torque. When the stick is pointing straight down, for example, the angular acceleration should be zero.

Thus, unless you were given that angular velocity when the stick is pointing down (or given a formula to find it), I think you'll have to calculate it by integrating the torque divided by the moment of inertia over 90 degrees of arc.

So exactly how I should get the angular speed for this question?

I wrote the comment wrong, I mean how I should get the acceleration of the meterstick when it is vertical?

When the meterstick is vertical, does it move?

That is, place it in the vertical position and let go.

As for the speed, consider the gravitational energy of the rule. The centre of gravity is at 0.25 m from the lowest point that it can reach.

Therefore, relative to that point, it has mgh potential energy.

When it swings down, all this energy is converted to kinetic energy, and you get the speed of the centre of gravity of the rule, v1.

The tip of the rule will undergo a greater change in speed, which is similar to the effect of moments.

v1d1 = v2d2

v2 = v1d1/d2

d1 = 0.25 m
d2 = 0.75 m

find v2, the velocity of the tip. We'll rename it v0 for the following part.

Using and , you get:



You have v0, g, l, so, you can get the velocity v of the tip of the ruler at any time t.

To get the angular velocity, simply use



Where r = 0.75 m

Nice job Jerry. It occurred to me after I went to bed last night that it would be easier to compute angular speed from conservation of energy than from integrating the torque.

As usual, it won't let me give you any rep. Congrats on breaking 3000 though! :)

Yes, I was thinking about that too and edited my original post.

Thanks, it were you who gave me my 3005th point ;)