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In my dynamics class, we're being asked to solve the following problem:

enter image description here


My Attempt:

Since I'm given the initial velocity, final velocity, and distance, I solved for the acceleration of the plane using kinematics:

$a = \frac{v_f^2 - v_i^2}{2d} = \frac{(55.6 m/s)^2 - (16.7 m/s)^2}{2(425m)} = 3.31 m/s^2$

Now this is where my approach starts to differ from the formulas recommended by the class. I decided to set the moment about point A equal to zero, because the plane is not rotating as it moves down the runway. I seem to remember from both statics and physics that if a body is not rotating about a given point, you can simply set the moment around that point to be zero.

Let N be the reaction force at B. Combining the above assumption with Newton's second law in the x direction, I get:

$\Sigma F_x: R = ma = (140000kg)(3.31 m/s^2) = 4.63 *10^5 N$
$\Sigma M_A: -(15m)N + (2.4m)mg - (1.8m)R = 0$

Solving for N gives me:

$N = 1.64*10^5 N$

According to the solution guide, this is incorrect.


Solution Guide's Explanation:

The guide uses a formula which was introduced to us in the textbook, which is as follows.

For some point P fixed on a rigid body with center of mass G, the moment about point P is given by:

$\Sigma M_P = I_G\alpha + ma_Gd$

where:
$I_G$ is the moment of inertia of the rigid body about G
$\alpha$ is the angular acceleration of the rigid body about G
$a_G$ is the acceleration of G
$d$ is the moment-arm distance, from P to G, of $ma_G$

The book cleverly chose a point C on the plane through which both R and A (the reaction force at wheel A) pass. See below:

enter image description here

Using the moment equation above and setting $\alpha = 0$ (because the plane isn't rotating) gives:

$\Sigma M_C = ma_Gd = (15m)N - (3m-1.8m)mg$

Finally, using $a_G = 3.3 m/s^2$ and solving for N, they got:

$N = 2.57*10^5 N$


Why I'm Confused:

From the physics and statics classes I've taken in the past, I've always been taught that $\Sigma M = I\alpha$; there was never that extra "$+ mad$" term thrown onto the end. That term is basically saying that an accelerating body with no rotation can still have a moment about a point. By contrast, in my physics and statics classes, I recall using the fact that if a body is stationary (no angular acceleration), we can set the moment about any point on the body equal to zero to help us solve.


My Guess at Where the Inconsistency Lies:

Here's my guess at where the inconsistency lies: in my statics classes, we assumed that the rigid bodies we were analyzing had not only zero angular acceleration, but also zero linear acceleration. (It's statics, after all!) In dynamics problems such as this one, however, there is a nonzero linear acceleration which must be accounted for. The "+mad" term must come from the fact that for any point object, the moment about some fixed point O is given by:

$\Sigma M_O = \vec{r} \times \vec{F} = \vec{r} \times m\vec{a}$

As I'm writing this, I think the intuition totally makes sense. It seems the formula used by the solution guide accounts for both rotation of the rigid body about its center, and linear acceleration of the body with respect to some point P outside the center of mass of the body.

I'll still post this though, in case anyone would like to correct me, add anything, or use this for their own reference.

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  • $\begingroup$ Your image font is not readable. Would you add the text of the image to your question. Thanks. $\endgroup$ – kamran Apr 30 at 0:42
  • $\begingroup$ This is actually for once such a nice homework question, so much information and own effort instead of just a bad photo of a problem sheet. +1, thanks OP $\endgroup$ – OpticalResonator Apr 30 at 8:00
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There is a slight mistake in your approach. It seems you have forgotten to add the contribution the moment due to inertial force acting at the center of mass $G$ to the moment equation. The moment equation should be : $$\sum M_A = -(15 m) N + (3 m) 140*10^3*a + (2.4 m) 140*10^3*g - (1.8 m) 140*10^3*a = 0$$ Evaluating for $N$ results in : $N = 2.57 * 10^5 N$

Remember, the total moment at any point for both dynamic and static equilibrium problem evaluates to zero always. Hope it helps you!

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  • $\begingroup$ For reference, this is called D'Alembert's Principle, which basically means you can treat dynamic problems as static, but only if you properly account for the inertial forces. $\endgroup$ – Wasabi May 4 at 16:44
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You can't set the moment to zero unless the body isn't constrained in its movement. If you could set any moment to zero that didn't move, statics wouldn't work.

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