I see. I changed all of the trig functions to the inverses, but it is still not working out, I can't seem to find a simpler way to relate the info I am given and write better equations.
I set these equations up and I tried solving on my ti-89 but it did not work. I think I am thinking too much into this because it should be a fairly easy question.
Any advice for a new plan of attack?
Homework Statement
I attached a picture of the question.
Homework Equations
∑Fx=0
∑Fy=0
100kg(9.8)= 981 N
I am not sure how to start this problem. Everything else in this chapter is a breeze so perhaps a small insight or how to start would be sufficient, thanks.
Yes that is what I mean. I figured the force at D is 20lb because the block has 20lb of force and the cable holding it has 20lb of tension so D would as well (same cable). So I tried to resolve that 20lbs into the x and y direction.
I drew a FBD about point A and came up with the two equations:
∑Fx=0 20sinθ - T-ab(cos70)=0
ΣFy =0 -20cosθ + T-ab(sin70) -20=0
My answer is not coming out so I either messed up on the FBD or the equations, some help would be appreciated. I'm on 3-10 by the way.
To find the momentum change for an object in 2 dimensions do I find the initial momentum in both the x & y direction then apply the Pythagorean theorem, and then do the same for the final momentum and then find the difference between the two?
Yes it is a totally inelastic collision, and the 498 is the speed of both after they become one. Could you elaborate more on this centre of momentum frame?
Homework Statement
Spaceship mass=450000kg Initial Velocity= 300m/s y + 400m/s z
Asteroid mass = 2000kg Initial Velocity= -100m/s y - 50m/s z
Speed after 498m/s
Question: If the force between the objects is 5000N, how long will it take for spaceship to capture the asteroid? (Assume...