Normal reaction force of a roller to a wedge and the velocity

[MechaMZ]

Homework Statement

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Cam mechanisms are used in many machines. For example, cams open and close the valves in your car engine to admit gasoline vapor to each cylinder and to allow the escape of exhaust. The principle is illustrated in Figure P5.66, showing a follower rod (also called a pushrod) of mass m resting on a wedge of mass M. The sliding wedge duplicates the function of a rotating eccentric disk on a camshaft in your car.. Assume that there is no friction between the wedge and the base, between the pushrod and the wedge, or between the rod and the guide through which it slides. When the wedge is pushed to the left by the force vector F , the rod moves upward and does something such as opening a valve. By varying the shape of the wedge, the motion of the follower rod could be made quite complex, but assume that the wedge makes a constant angle of θ = 15.0°. Suppose you want the wedge and the rod to start from rest and move with constant acceleration, with the rod moving upward 1.00 mm in 8.00 ms. Take m = 0.250 kg and M = 0.650 kg. What force F must be applied to the wedge?

The Attempt at a Solution

Below are the correct working steps:

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y_f = y_i + V_yit + 0.5a_yt²
(1x10^-3) = 0 + 0.5ay(8x10^-3)²
a_y = 31.25

I don't understand why I couldn't get the acceleration by the method below:
V_m = 0.125m/s since the wedge and the rod to start from rest and move with constant acceleration, with the rod moving upward 1.00 mm in 8.00 ms

Then calculate out the V_M by using the Pythagorean theorem, VM = 0.466m/s
Then using V_xf = V_xi + a_xt to find out the ax
V_xf = V_xi + a_xt
0.466 = 0 + a_x(1s)
a_x = 0.466m/s²

tan15 = y/x
a_x = 116.626m/s²

Sum F_y = ma_y
ncos15 - 0.250kg(9.81) = 0.250kg(31.25)
n = 10.627

Why the normal reaction force, n is not equal to the (0.25g)cos15, but using the method above to find it?

Sum F_x = max
-nsin15 + F = 0.65(116.626)
-10.627sin15 + F = 0.65(116.626)
F = 78.557N

Hope somebody could explain this to me =)

 

[method_man]

V_m = 0.125m/s since the wedge and the rod to start from rest and move with constant acceleration, with the rod moving upward 1.00 mm in 8.00 ms

Should this be rod's velocity? How can you determine velocity if it accelerates constantly?