# forces problem

• Jul 2nd 2007, 01:37 PM
forces problem
hello , i would like you please to find the solution for all of those ,am find it difficult to do them alone i need help!!:eek:
• Jul 2nd 2007, 06:50 PM
topsquark
Quote:

hello , i would like you please to find the solution for all of those ,am find it difficult to do them alone i need help!!:eek:

For #1

Recall Newton's 2nd Law:
$\sum F = ma$

and the rotational version of Newton's 2nd Law:
$\sum \tau = I \alpha$

Since the plank is in static equilibrium $\sum F = 0$ and $\tau = 0$

So choose a +y direction to be directly upward. A Free Body Diagram shows that there are four forces present: The forces A and B acting directly upward, the weight w of the woman acting directly downward, and the weight W of the plank also acting directly downward. So we know that
$\sum F_y = A + B - w - W = 0$

$A + B - mg - Mg = 0$
where m and M are the mass of the woman and the plank respectively.

Also choose a positive rotation to be in the counterclockwise sense. I am going to choose an "axis of rotation" to be at the point where the reaction force A is operating. (We may choose any point as our axis, since there is no rotation anyway.) We don't know where the CM of the plank is yet, so I'm going to place that at a distance x to the right of the axis of rotation. So
$\sum \tau _A = -(2)w - (x)W + (6)B = 0$<-- That's "2 meters there as the coefficient of w, etc.

$-2mg - xMg + 6B = 0$

We have three unknowns in these two equations.

Well, pick a new axis of rotation and do it again! :) I'll now pick the axis at the point where B is acting, with the same rotation convention. So
$\sum \tau_B = (6 - x)W + (4~m)w - (6~m)A = 0$

$(6 - x)Mg + 4mg - 6A = 0$

This gives us the system of equations:
$A + B - mg - Mg = 0$
$-2mg - xMg + 6B = 0$
$(6 - x)Mg + 4mg - 6A = 0$

We have three equations in three unknowns (A, B, and x), so we may solve this.

I have to get going. If you have a problem solving the system just post in the thread and someone will help you.

-Dan