- #1

gnits

- 137

- 46

- Homework Statement
- To find the angle between the ground and rods when in limiting friction

- Relevant Equations
- force balancing and moments

Could I please ask for help with the following:

Here's my diagram:

The forces at the hinge (green) are internal forces.

For the whole system resolving vertically gives:

R1 + R2 = 4W

and horizontally gives:

F1 = F2

For the rod of weight 3W only, taking moments about B gives:

F1 * L * sin(Ѳ) + 3W * (1/2) * L * cos(Ѳ) = R1 * L * cos(Ѳ)

When in the limiting situation I can replace F1 with (2/3) * R1 and rearrange to give:

(2/3) * tan(Ѳ) = 1 - ( 3W / (2 * R1) )

For the rod of weight W only, taking moments about B gives:

R2 * L * cos(Ѳ) = F2 * L * sin(Ѳ) + W * (L/2) * cos(Ѳ)

When in the limiting situation I can replace F2 with (2/3) * R2 and rearrange to give:

(2/3) * tan(Ѳ) = 1 - ( W / (2 * R2) )

Now, and maybe this is bad reasoning?, we always have that F1 = F2, so in the limiting situation when either F1 = (2/3) * R1 or F2 = (2/3) * R2 then we will have that R1 = R2.

So, from the above equations, replacing R2 with R1, I will have:

For rod of weight 3W:

(2/3) * tan(Ѳ) = 1 - ( 3W / (2 * R1) )

and for rod of weight W:

(2/3) * tan(Ѳ) = 1 - ( W / (2 * R1) )

And the latter is the larger value and so the latter has the larger value of Ѳ in the limiting case and so rod BC will be the one which slips.

This agrees with the book answer.

However, for the next part, determining the angle, I do not get the book answer of 45 degrees.

Here I equate my two equations to give:

1 - (3*W) / (2 * R1) = 1 - W / (2 * R2)

which gives:

R1 = 3 * R2

and using R1 = 4W - R2 leads to R2 = W

Which, if put back into the expression for (2/3) tan(Ѳ) above, leads to tan(Ѳ) = 3/4

which does not imply that Ѳ = 45 degrees.

Where did I go wrong?

Thanks for any help.

Here's my diagram:

The forces at the hinge (green) are internal forces.

For the whole system resolving vertically gives:

R1 + R2 = 4W

and horizontally gives:

F1 = F2

For the rod of weight 3W only, taking moments about B gives:

F1 * L * sin(Ѳ) + 3W * (1/2) * L * cos(Ѳ) = R1 * L * cos(Ѳ)

When in the limiting situation I can replace F1 with (2/3) * R1 and rearrange to give:

(2/3) * tan(Ѳ) = 1 - ( 3W / (2 * R1) )

For the rod of weight W only, taking moments about B gives:

R2 * L * cos(Ѳ) = F2 * L * sin(Ѳ) + W * (L/2) * cos(Ѳ)

When in the limiting situation I can replace F2 with (2/3) * R2 and rearrange to give:

(2/3) * tan(Ѳ) = 1 - ( W / (2 * R2) )

Now, and maybe this is bad reasoning?, we always have that F1 = F2, so in the limiting situation when either F1 = (2/3) * R1 or F2 = (2/3) * R2 then we will have that R1 = R2.

So, from the above equations, replacing R2 with R1, I will have:

For rod of weight 3W:

(2/3) * tan(Ѳ) = 1 - ( 3W / (2 * R1) )

and for rod of weight W:

(2/3) * tan(Ѳ) = 1 - ( W / (2 * R1) )

And the latter is the larger value and so the latter has the larger value of Ѳ in the limiting case and so rod BC will be the one which slips.

This agrees with the book answer.

However, for the next part, determining the angle, I do not get the book answer of 45 degrees.

Here I equate my two equations to give:

1 - (3*W) / (2 * R1) = 1 - W / (2 * R2)

which gives:

R1 = 3 * R2

and using R1 = 4W - R2 leads to R2 = W

Which, if put back into the expression for (2/3) tan(Ѳ) above, leads to tan(Ѳ) = 3/4

which does not imply that Ѳ = 45 degrees.

Where did I go wrong?

Thanks for any help.