Help Solving 2 Qs: Tan(Ɵ) & Sign Direction Error

In summary, the back of the book says that if you equate forces vertically for the vertical rod only you get:F1 = W/2If you equate forces for the whole system vertically you get:R = (3/2)WHowever, if you take clockwise moments about C for rod CD only you get:F * L sin(t) + W * (L/2) cos(t) = R * L cos(t)Substituting for R and rearranging gives: F = W / tan(t)
  • #1
gnits
137
46
Homework Statement
To find possible angle of two leaning rods
Relevant Equations
moments
Could I please ask for help with the following question:

1642514927301.png


The last part follows easily from the first part.

Answer from back of book for first part is:

2/(3u') <= tan(Ɵ) <= 2u

What I have done is the following:

Here's my diagram (I have separated the components to show the internal forces in the system. I have used t instead of Ɵ) :

leaning.png


Orange forces are internal forces.

If I equate forces for the whole system vertically I get:

R = (3/2)W

If I equate forces vertically for the vertical rod only I get:

F1 = W/2

If I take clockwise moments about C for rod CD only I get:

F * L sin(t) + W * (L/2) cos(t) = R * L cos(t)

Substituting for R and rearranging gives: F = W / tan(t)

Now for no slipping at D we need F <= u' R so this gives:

W/tan(t) <= u' * (3/2)W

Which leads to tan(t) <= 2/(3u') which is the answer asked for but with the sign reversed.

1) How have I gotten the sign mixed up?

Taking clockwise moments about D for rod CD only I get:

S * L sin(t) = F1 * L cos(t) + W * (L/2) * cos*(t)

Substituting for F1and rearranging gives:

S = W/tan(t) and for no slipping at C we need S <= u * F1

So gives W/tan(t) <= u * W/2 which leads to tan(t) >= 2/u

Again wrong direction of sign and also wrong answer as we need tan(t) <= 2u

2) Where have a erred?

Thanks for any help.
 
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  • #2
Thank you very much kuruman for your reply.

So, notwithstanding the signs, would you (or others) think my analysis of taking moments for CD only about D looks correct? I ask because here I differ in the actual form of the answer. The given answer is 2u but I arrive at 2/u. If you\others agree that 2/u is correct then I would put it down to a typo in the book.

Thanks again.
 
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  • #3
gnits said:
Now for no slipping at D we need F <= u' R so this gives:

W/tan(t) <= u' * (3/2)W

Which leads to tan(t) <= 2/(3u') which is the answer asked for but with the sign reversed.
I agree with the first equation here, but that implies that ##\mu' \tan \theta \ge \dfrac 2 3##.
 
  • #4
PeroK said:
I agree with the first equation here, but that implies that ##\mu' \tan \theta \ge \dfrac 2 3##.
Yes indeed, I agree. I had fallen at the last there. Thanks for that. So only the second part is now causing me issues.
 
  • #5
gnits said:
S = W/tan(t) and for no slipping at C we need S <= u * F1
For no slipping we need ##\mu S \ge F_1 = \dfrac{mg}{2}##.

##S## is the normal force and ##F_1## is the required friction force.
 
  • #6
PeroK said:
For no slipping we need ##\mu S \ge F_1 = \dfrac{mg}{2}##.

##S## is the normal force and ##F_1## is the required friction force.
Thanks very much indeed. I had swapped S and F1. I had written S <= u * F1 whereas, as you say, I should have written F1 <= u S. And indeed this leads to the book's answer. Thank you very much.
 
  • #7
gnits said:
Thank you very much kuruman for your reply.

So, notwithstanding the signs, would you (or others) think my analysis of taking moments for CD only about D looks correct? I ask because here I differ in the actual form of the answer. The given answer is 2u but I arrive at 2/u. If you\others agree that 2/u is correct then I would put it down to a typo in the book.

Thanks again.
Sorry, I deleted my reply to write a better one, but then I had an emergency that kept me away from the task. You must have read my post soon after I posted it but before I deleted it. Anyway, it looks like your issue has been resolved so there is nothing more for me to say. I apologize for the confusion.
 
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