- FBD (as shown)
- Newton-Euler: Sum forces in x- and y-directions. [With choice of xyz axes shown, N_A appears in only the y-equation which simplifies analysis somewhat.] Set forces equal to m*a_G. Suggest using moment (Euler) equation about G (cm of bar). These give you three equations in terms of 4 unknowns (a_Gx, a_Gy, alpha and N_A).
- Kinematics: Need at least one additional equation. Suggest relating the acceleration of G to the acceleration of A (you know direction of a_A). This gives you two scalar equations (x- and y-components) with an additional unknown (a_A) for a total of five equations in terms of five unknowns.
- Solve: Solve the above equations for a_A.
Note that you could replace the above Euler equation about G with an Euler equation about A (this would eliminate N_A). However, be careful. By using A, you will need the following terms on the right hand side of the Euler equation: I_A*
alpha + m*(
r_G/A) x (
a_A).
7 comments:
is the sum of F in the y direction 0 or does N become something other than mg*cos(theta)?
The RHS of the summation of forces in the y-direction is equal to m*a_Gy, which is NOT zero.
a_Gy is one of the five unknowns in the problem solution.
Is there a trigonometry trick that needs to be used to get the final answer? I got a_A=-g*sin(theta)-[3*g*cos(theta)^2]/[1+3*sin(theta)^2]. I have five equations, three from Newton, two from kinematics and the rigid body equation between G and A. The equations I plugged into my a_A equation (x direction only) : a_Gx=-g*sin(theta) Fn=m*g*cos(theta)/[1+3*sin(theta)^2] Any hints?
Two suggestions that come to mind:
* Put your two terms over a common denominator to get a single fraction.
* Use the trig identity of sin(theta)^2 = 1 - cos(theta)^2
This will give you the correct denominator. I will also work through to see if this gives you the correct numerator.
Let me know if this works for you.
I get the same expression as you do for Fn (the normal force at A).
I recommend that you use the two suggestions that I made above and simplify. If you do not get the correct answer, then recheck your algebra.
Sigh... I spent so long trying to figure out why my answers weren't coming out right, I was off by one little coefficient in the denominator, and it all came down to how I defined my axes. I figured it doesn't matter which way you point x as long as you keep signs straight, but I guess when you make x point to the left and leave k coming out of the page, things get nasty. Hope no one else is too stubborn to use the axes already given in the picture.
Post a Comment