Recent content by AntSC

But the acceleration function is a piecewise function and the ##\frac{1}{4}t^2## function you have used only applies in the ##1\leq t\leq 6## domain and not ##0\leq t\leq 6##. I did test this with integration too and still got ##10ms^{-1}## but maybe check again if you have the inclination. I...

There is a thread on the use of Latex for PF if needed.

Looks good to me. Yes, it was your rearrangement of ##F=mg\tan \theta## that was the issue. Everything else seems good

Thanks BvU! You've helped my sanity. It's always uncertain with answers for multiple-choice questions as there's no working, so you can't retrace any steps. Glad there's some agreement here

Error in your first equation. Otherwise, all the components are there and looks good

The variation with time t of the acceleration a of an object is shown What is the change in velocity of the object from ##t=0## to ##t=6##? A. ##6ms^{-1}## B. ##8ms^{-1}## C. ##10ms^{-1}## D. ##14ms^{-1}## So apparently the answer is B, which I am having trouble reconciling. Using methods...

I understand that the kinematic equations work for constant acceleration but they're derived from first principles by integrating with respect to time. I don't see how replacing ##a## for a function of height would have any effect on the derivations. But should we rederive suitable equations by...

I recently watched a video on youtube where a guy fires a 50 cal. rifle straight up into the air and measures the time of flight at about 100s. It got me thinking about what altitude the round reached. So I used the kinematics equations and obtained a value of 12.25km. Amazing, but I wondered...

Hi Haruspex. Taken me ages to reply. Busy time of year! Following off from your last message, if we assume the main rod is perpendicular to the wall but the support rod is not perpendicular to the main rod, how would I accurately consider all the forces acting on the main rod if I was going to...

Thanks for pulling me up on my considerations. I meant to only point out that I just wanted to know what happens at the point of support. I guess I also didn't mention that this is a statics problem. If so, does that still make the weight of the horizontal rod and the forces where it the...

In which direction does the support force act on the horizontal rod? All the problems I'm looking at are giving correct answers only if I assume the support rod is applying a force to the horizontal rod in a direction parallel to the support rod.

Homework Statement This is more of a question of applying the correct principles to a range of problems involving a horizontal rod hinged to a wall that is then supported by another rod. The support rod is also attached to the wall from above or below and then attached to the horizontal rod, so...

Not sure whether that's a sarcastic remark or not. Thanks for spotting my lame mistake. Must have repeated it each time. It all works out now. I got all 3 coefficients. Perhaps we could talk further about the logic behind setting these x-values. I understand your point of not allowing x=-2/5 and...

From what i can see from textbooks the method I've chosen is valid. The reason behind making x=-2/5 and x=3/2 is to eliminate certain coefficients, leaving one unknown in each case thus making each coefficient solvable. Not quite sure of the logic of not dividing by those values, please explain...

Homework Statement The question is stated at the top of the attached picture with a solution 20160303_095831.jpg The correct results of the coefficients are A=2, B=-5, C=1 I have tried this problem multiple times and am still getting ugly coefficients. I have no idea why. A fresh pair of eyes...