How far up a ladder could you climb before it slips?

[@ardordeleon]

10:00 You wanted to divide everything by cos(theta) and then divided the last term on the RHS by sin(theta). You got an extra 1/tan(theta) that shouldn't be there

[@fizixx]

Yes, I was going to mention that as well.

[@fizixx]

This is a great problem because it involves many things you don't usually see in these kinds of problems, and the math isn't really all that involved.

[@spencergee6948]

Yes, there should not be a second tan (theta). Perhaps a better way is to arrange the final (correct) equation for tan(theta) and see, for example at what angle the ladder would slip for various values of s/l. For s/l = 0.5 (i.e. half way up the ladder) then tan (theta) = 1/2u . If friction at the wall (u2) is included, the equivalent equation for tan (theta) becomes tan (theta) = (1+u1.u2)/2u1. Clearly, if u2 = 0 we have the original relationship.

[@brookestephen]

I must admit I was hoping for some calculus, and some Python programming to visualize the slide of the ladder! BTW you're dividing both sides by Mh cos Theta... not Mh sin Theta... check the term on the far-right. NOW you might want to restart the video!

[@boltvalley3076]

Very smart 🤓

[@randymeyer6482]

Get an HP RPN calculator..an 11C or 32ii...

[@lisazavieh78]

Unfortunately, I think the answer is incorrect - it should be 4.3 m, so the ladder doesn't slip at all. Summing torques around the bottom contact point, I get 0 = (750)(cos60)(x) + (50)(cos60)(2) - (480)(sin60)(4) ; x = 4.3 m.

[@Roo1717]

i got 4.3 too. seemed to be overcomplicated? I don't follow why they did N2*sin(theta) in the moment sum. where is the moment arm? with the given angle and 4m ladder, the moment arm for N2 should simply be 3.46 m, meaning we already know N1, N2, and Ff. My torque sum was mLgcos60(L/2) + mHgcos60(s) - N2(3.46) = 0 ; s = 4.3 m.

[@hoyeeyeung6938]

Same, got 4.3m