Solving the Slender Ladder Problem: Angular Acceleration

[eurekameh]

The slender ladder has a mass of 10 kg. It is released from rest in the position shown. Friction at the two contact surfaces are negligible. Determine the angular acceleration of the ladder.

I summed forces in the x,y direction, summed moments about the ladder's center of mass, wrote a kinematics equation relating the angular acceleration to the acceleration of its center of mass for a total of 5 unknowns (Force at bottom, ax, ay, angular acceleration, force at the vertical wall) and 5 equations. I have angular acceleration = 3.75 rad/s^2. This was on a multiple choice test, but there was no answer of 3.75 rad/s^2. I've been at this problem for days, making the same "mistake" over and over. I have found that the solutions manual of a similar problem is also doing the same thing I'm doing. Did my professor make a mistake?

 

[tiny-tim]

hi eurekameh!

(pleeeease don't post such wide images )

(it should be possible with just one τ = Iα equation, and one constraint equation … anyway:)

show us your full calculations, and then we'll see what went wrong, and we'll know how to help!

 

[Quinzio]

3.75 rad/s^2 is correct.

 

[eurekameh]

tiny-tim:
I don't think it's possible with just one tau = (I)(alpha) equation and one constraint equation because then you'd have too many unknowns with not enough equations.
Denoting Fx as the force from wall, and Fy as the force from floor:
For forces in x;
Fx = max
For forces in y;
Fy - mg = may
And for moments about center of mass;
(Fy)(cos60) - (Fx)(sin60) = (1/12)(m)(l^2)(alpha)
The two constraint equations gave me:
ay = -(alpha)(cos60)
ax = (alpha)(sin60)
5 unknowns in 5 equations gave me 3.75 rad/s^2.
Thanks for the confirmation, Quinzio. :)

 

[tiny-tim]

hi eurekameh!

tiny-tim:
I don't think it's possible with just one tau = (I)(alpha) equation and one constraint equation because then you'd have too many unknowns with not enough equations.

yes you can if you do it τ = dL/dt about P, the point where the normals meet …

τ = dL/dt = d/dt {I_c.o.mω + mr_c.o.m x v_c.o.m} = I_c.o.mα + mr_c.o.m x a_c.o.m …

give it a try! ​

(i don't know why i said Iα before )

 

[sk9]

YES, so it by instantaneous axis of rotation. search it up on google if you don't know about it. it becomes very easy