Recent content by kspabo

Let's say I have a 15-lb flag that is mounted to a car on a welded bracket. I want to know the general force the flag will exert on the bracket (neglecting flex in the flag, assuming uniform weight throughout) in the scenario of let's say braking at max speed (as this is the most simple...

Homework Statement Homework Equations Coriolis Accel = 2*v*w Where w is the angular velocity and v is the linear velocity. The Attempt at a Solution I really just need help on understanding how the Coriolis acceleration in this problem is depicted. My dynamics book is extremely confusing...

Thank you andrewkirk, that makes sense.

Homework Statement Homework Equations Skipping this, as I just have a simple question The Attempt at a Solution Here is the full solution, but I just have one simple question about a simple step. On the part for disk B, they establish the equation vD = vB - vD/B. In the proceeding step...

My apologies, I completely forgot to include that part. Here it is:

Homework Statement Homework Equations [/B] Radial and Transverse coordinates to relate acceleration Balancing forces The Attempt at a Solution I know that there is more to this problem, but my question is simple: Why don't we know what the normal force is? When I solved it myself I...

That really clears it up. Thank you, I appreciate that.

Homework Statement Boxes are placed on a slope at uniform intervals of time trelease and slide down the slope with uniform acceleration. Knowing that as a box B is released, the preceding box A has slid 6 meters down the slope and that 1 second later they are 10 meters apart, determine the...

Wow I'm an idiot hah. For some reason I was doing the conversion wrong the 3 times I redid the problem. Thank you!

Homework Statement Two charged beads are on the plastic ring in Figure (a). Bead 2, which is not shown, is fixed in place on the ring, which has radius R = 63.3 cm. Bead 1 is initially on the x axis at angle θ = 0o. It is then moved to the opposite side, at angle θ = 180o, through the first and...

k q1/(2+x)^2 = -(k q2/x^2) q1/(2+x)^2 = -(q2/x^2) +9/(2+x)^2 = -(-1/x^2) 9/(2+x)^2 = 1/x^2 x^2/(2+x)^2 = 1/9 x/(2+x) = ± 1/3 (2+x)/x = ± 3 2/x + 1 = ± 3 2/x = -1 ± 3 So we have x=1 when -1 + 3 Ah, that makes sense and I believe is the correct answer.

Correct. So the positive field flows right towards the negative which continues a field to the right. I'm looking for the neutral point nearest the negative charge. How can I use this info to solve this if q1 is at 0 and q2 is at x=2?

Homework Statement Two charges q1=+9c and q2+-1c are separated by 2m. Where is their neutral point? Homework Equations E = E(+) + E(-) E= k q/r^2 The Attempt at a Solution k q1/(2+x)^2 = -(k q2/x^2) When I solve for this equation I end up square rooting a negative number which would then...

EDIT: My work was wrong and this post was pointless. Currently working it out again. EDIT2: I finally got the answer. I'll post my work for future people that have this same question. sqrt(3)/3 = r/(L-r) L sqrt(3)/3 - r sqrt(3)/3 = r L sqrt(3)/3 = r (1+ sqrt(3)/3) r = (L sqrt(3)/3)/(1+ sqrt(3)/3)

Oh wow, I see what you're talking about. I blame my long summer break for this silly arithmetic error. Ok, so I fixed my work and have arrived at: q1/r2 = 3q1/(L-r)2 1/r2 = 3/(L-r)2 sqrt(1/3) = sqrt(r2/(L-r)2) sqrt(1/3) = r/(L-r) L(sqrt(1/3) + 1)= r So after going through this I just realized...