- #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:

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) :

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.

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

Thanks for any help.

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) :

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.