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 GQ311 chale21 Thu 12/2 20:27 How do you know which direction to draw pin reaction forces on FBDs? I keep having trouble with the direction of them being wrong. 0 kkit Thu 12/2 21:33 Draw them in the +x and +y directions. When you solve for the forces, their sign will tell you which direction they actually act in. 0 Add a response... GQ310 chale21 Wed 12/1 21:41 When doing some problems with projectile motion (for example 2014 practice test question 5), I was running into some problems with the constant acceleration equations. There are 4 different y-motion equations for constant acceleration in projectile motion, and to me, it looked like 3 could be used to solve the time as they all include the same variables. However, when solving for time, only one of them gave the right answer. Is there some way to know which one to use or should they all work and I'm just doing something wrong? Thank you 0 kkit Thu 12/2 9:22 There are multiple ways to solve the problem, but they will all give the same answer if done correctly. Can you explain which equations you are referring to? 0 chale21 Thu 12/2 15:49 Yea so for part c of question 5 on 2014 practice final exam, the constant acceleration equation used in the solution is y = y(sub 0) + V(sub y0)*t + 1/2*a(sub y)*t^2. However, I tried the equation V(sub y) = V(sub y0) + a(sub y)*t, using 0 as V(sub y), -16.94 as V(sub y0), and 32.2 as a(sub y) to solve for time, but it gave me a time of 0.53 seconds, which is incorrect. 0 nhicks7 Thu 12/2 15:56 So, the moment after the baseball is hit by the hail, it has the downward velocity of 16.94 ft/s. You gave this a negative value, suggesting up is positive and down is negative. So what should be the sign of ay? Should the baseball be speeding up as it falls to the ground or should it be slowing down to a velocity of 0? 0 nhicks7 Thu 12/2 16:01 The problem with using the equation you've indicated is that with that equation you have two unknowns. You know neither the time it will take the ball to reach the ground, nor the velocity with which it will hit the ground. It could still be used, but it would have to be used in conjunction with another equation or other equations to deal with that problem. Meanwhile, the y = y0 + vy,0t - 1/2gt2 equation has only one unknown variable: t. You know y0 (the initial height where the ball was hit by the hail), you know vy,0 (the downward velocity just after impact with the hail), you know g, and you know y (the final height, which is the height of the ground). 0 chale21 Thu 12/2 16:03 Oops sorry, I made a mistake typing. I had a(sub y) as -32.2 but I accidently typed it as positive. So is my problem that V(sub y) should not be equal to zero? I thought that we could consider the ball to have 0 velocity at the end since it is hitting the ground, but I now am thinking this is an incorrect assumption. 0 nhicks7 Thu 12/2 16:06 Right, because the only reason it would end up with a velocity of 0 is because the ground applies a force to it during a collision, which is no longer the constant acceleration context in which those kinematic equations apply. So you have to find it for the moment just before it contacts the ground. Knowing anything beyond that would require you to know the coefficient of restitution between the ball and the ground, because it's likely the ball would bounce back up after that, anyway, unless it were just a ball of putty or something else that would splat (and to be clear, even if that case, you still couldn't say vy = 0, because, again, there is a variable force that's applied in the collision with the ground to make that happen). 0 chale21 Thu 12/2 16:08 Ooh ok I get it now. Thank you very much! 0 Add a response... GQ309 chale21 Wed 12/1 18:42 On the 2014 final practice exam, on question 3 part A, I'm confused why the solution only has total work being the work of friction. Should gravity not be doing work as well as the block moves up the curve? 0 kkit Wed 12/1 19:47 The work of the weight is accounted for by the potential energy of the weight. The datum was chosen at point A so mgh_A = 0 and the mgh_c term on the right accounts for the work of the weight in moving from A to C. 0 Add a response...