At the instant when theta=30 degrees the horizontal guide is given a constant up
ID: 2144261 • Letter: A
Question
At the instant when theta=30 degrees the horizontal guide is given a constant upward velocity v0=2 m/s. For the instant calculate the force N exerted by the fixed circular slot and the force P exerted by the horizontal slot on the 0.5 kg pin A. The width of the slots is slightly greater than the diameter of the pin and friction is negligible.
I'm using rectangular coordinates to solve this problem. Summation of the forces in the X direction: I have Ncos(theta) = m(a_x). For the summation of the forces in the Y direction: I have P - W - Nsin(theta) = m(a_y), but a_y is zero, because the v_0 is constant in the vertical direction.
I am having a problem with finding the a_x (acceleration in the x direction?) Or should I be looking at this question in a different manner?
Explanation / Answer
You have the equation right for forces in the x direction.
You know that a_x = a_r/cos(theta),
a_r = v^2/r = ((2/cos(30))^2)/0.25 = 21.33 m/s^2
so a_x = 21.33/cos(30) = 24.64 m/s^2
using your equation for forces in the x-direction you get N = 14.22222 Newtons
You have some sign errors to solve for P, it should be P+W = Nsin(Theta) which gets you a value of 2.21 N for P
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