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?

 

[tomcue]

I would like to provide a response to the above content. The normal reaction force of a roller to a wedge is an important concept in understanding the motion of objects in cam mechanisms. In this particular scenario, the normal reaction force is the force exerted by the wedge on the rod, perpendicular to the surface of the wedge. It is not equal to the weight of the rod, which is what you calculated using the method of (0.25g)cos15.

The reason for this is that the normal reaction force is dependent on the motion of the rod and the wedge. In this case, the rod is moving with a constant acceleration, which means that there is an unbalanced force acting on it. This unbalanced force is the normal reaction force, which is equal to the mass of the rod multiplied by its acceleration. Therefore, the normal reaction force in this scenario is not equal to the weight of the rod, but rather to the product of its mass and acceleration.

In order to find the normal reaction force, we need to use the equation F = ma, where F is the normal reaction force, m is the mass of the rod, and a is its acceleration. Using this equation, we can solve for the normal reaction force and find that it is equal to 10.627N. This is the same value that you obtained using the method above, which is the correct approach.

In summary, the normal reaction force is not equal to the weight of the rod in this scenario because the rod is moving with a constant acceleration, which means that there is an unbalanced force acting on it. This unbalanced force is the normal reaction force and is equal to the product of the mass and acceleration of the rod. I hope this explanation helps to clarify any confusion you may have had.