Recent content by kuruman

I am adding this note to complete your understanding. The expression you derived for the final velocity of the coalesced object is the velocity of the center of mass (CM) which remains unchanged throughout the collision because there are no external forces acting on the combined bullet plus...

Total momentum is always conserved in a collision. The point of my post was to show that all the energy of the "bullet" cannot be transferred to the target as claimed because addition of the necessary condition that momentum is conserved constrains the masses to be equal. In other words, because...

To begin with, this problem is absurdly unrealistic. A 123 kg bullet? The mass of a common 55 mm artillery shell is about 43 kg. Setting that aside, we can certainly write down the energy of the bullet as ##K_b=\frac{1}{2}m_bv_0^2##. The question is, how much of that is transferred to the...

Please explain what you mean by ##\theta## when you have a Γ - shaped construction. It's the angle between what two lines? It seems to me that you replace the original situation in post #1, which has a well-defined angle ##\theta##, with a different situation shown in post #6, where ##\theta##...

When you set ##\theta=0## (or ##\theta=\frac{1}{2}\pi##) you are effectively saying that there is no diagonal piece, just a vertical piece and a horizontal piece screwed to the vertical piece and the contraption looks like the letter Γ . That is problematic because the screw/glue or whatever...

Whether it is correct or incorrect, it cannot help analyze the system because you have not defined what the system is. Defining the system sets the boundary between the system and the external world thereby determining the forces that you put in the diagram. Any entity outside this boundary may...

Not quite. The equation says that you need an infinite force to open a door if you push at the hinge. When the location of the hinge is not obvious, to prevent people from pushing at the wrong place (and to avoid unnecessary cussing), doors in public places often have PUSH signs to ensure that a...

When I was active teaching I talked turkey to my students about this on the first day. This is what I said. I cannot guarantee that you will learn anything in this class because if you don't want to learn, you will not. However, I can guarantee that if you do want to learn, you will. Think...

I agree with @A.T. 's suggestion

The symbols labeling the arrows in FBD represent magnitudes. One has to pick a direction for the tip of the arrow often without knowing beforehand if the direction is correct. This is necessary for writing down the equations and doing the algebra to get expressions for the unknown quantities...

Your equation ##~L_2w=hF_x~## in #16 gives $$F_x=\frac{L_2}{h}W.$$ If you cast it in terms of the given quantities, namely ##L_1## (0.16 m) and ##\theta## (55o), you would substitute ##~h=\dfrac{L_1}{\tan\theta}~## to get $$F_x=W\left(\dfrac{L_2}{L_1}\right)\tan\theta.$$That is exactly my...

It seems you missed what post #13 shows. I chose as system the horizontal bar only. The physical situation says that the items exerting external forces on this system are The wall at the point of contact which could be a screw. The diagonal piece pushing from underneath up and to the right...

This is not the same as this which is the figure found in the original post. The answers using the free body diagram and the equations in post #13 are

Having the net force equal to zero also matters, of course. The forces acting on the horizontal bar arise from three external entities, the vertical piece (or wall), the diagonal piece and the Earth. There can be no couple because with three forces, there is always an odd one out. However, if...

Yes, of course. The horizontal bar doesn't give a hoot about what we choose as "the system" as long as the "gallows" configuration remains unchanged. If the diagonal piece just touched the wall without being fastened and the contact were frictionless, then ##F_y## would be equal to the hanging...