Solving the Overturning Moment Problem: 4 Seconds

[a150daysflood]

Homework Statement

In this problem, take g = 10 m s-2.
A uniform plank of length 14 m long and mass 20 kg is being supported at B and C as shown above.
Two boys X and Y of masses 20 kg and 50 kg respectively are right on top of the supports.
X starts to walk towards A and Y walks towards D at the same time.
If both boys walk at 1.0 m s-1, the plank begins to overturn after how many seconds?

Homework Equations

Clockwise moment = anticlockwise moment
fs=fs

The Attempt at a Solution

The answer given is 4seconds.
i can't work it out,really need help.
Need a little hint please.
Thanks.

 

[brett351]

Imagine support C is not there. Imagine each boy has walked for 1 second. Each boy exerts a moment around B equal to is weight times his distance from B. If BoyY's moment is larger, the plank will want to rotate CW. But the plank doesn't rotate because support C is there. Will BoyX's moment eventually become larger? If so when?

Now do the same thing imagining that support B is not there.

 

[Kenny Lee]

Sup flood, this is what I got:

F_y equilibrium gives: B + C = 90g
Moment equilibrium about B gives: t = (C - 60g)/ 7.5g,

Where B and C are the reactions at B and C, and t is the time after X and Y start moving.

Recall that the reactions are given by:

B + C = 90g

I figured that since Y is heavier, the beam will overturn when all the weights are centered at C i.e. when B = 0 and C = 90. Subbing this into the moment eqbm equation, we get:

t = 30g/ 7.5g = 4 s

Which is the answer yea? How'd you get 1 s? Anyhows... yea let me know if that helped.

...
Kenny

 

[a150daysflood]

Yea kenny lee your right.
Thanks a lot.

 

[Kenny Lee]

Glad to be of help =)

 

[a150daysflood]

But how did you get 7.5g anyway?

 

[Kenny Lee]

I've attached my working. Its Jpeg. Um you have a look, see if it makes sense. (or if its even right).

G luck

 

[a150daysflood]

Thanks a zillion for you effort and time.
God bless..