Two sticks are connected, with hinges, to each other and to a wall. The bottom stick is horizontal and has length \(L\), and the sticks make an angle of \(\theta\) with each other, as shown in Figure above. If both sticks have the same mass per unit length, \(\rho\), find the horizontal and vertical components of the force that the wall exerts on the top hinge.

Also show that the magnitude goes to infinity for both \(\theta\)→ \(0\) and \(\theta\)→\(\frac{\pi}{2}\)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestalt text

In the above figure, I have marked all the forces that are present in the system. Let the masses of the two rods be \(\displaystyle m_1\) and \(\displaystyle m_2\).

Thus,

\[m_1 = \rho L\sec\theta\] \[m_2 = \rho L\]

Considering the torque about the point A,

\[h_1(L\tan\theta) = (m_1+m_2)g\frac{L}{2}\]

\[h_1(L\tan\theta) = \rho L (1+\sec\theta)g\frac{L}{2}\]

\[\boxed{h_1 = \rho\frac{1+\sec\theta}{\tan\theta}\frac{gL}{2}}\]

Ill do the vertical force later..Ill post it as a comment

Log in to reply

Why are there no forces at the hinge joining the 2 sticks ??

Log in to reply

There are forces at the hinge joining the sticks, but they are internal forces. So, they cancel each other off..

Log in to reply

Log in to reply

Nice explanation

Log in to reply

The vertical force can be found using the pivot point B. It is -(rho

lg/2)(1+2sec(theta)).Log in to reply

do you know any good book with exersises in rotational dynamics ;

Log in to reply

Try IE Irodov or Krotov

Log in to reply