Recent content by ohphysics

I would do the tan inverse of that if it was the horizontal and vertical components. I can tell the cannon ball would be affected differently in the sub's reference frame because the submarine is going the opposite direction but I'm not sure how to quantify that

horizontal component is 60cos 45 and vertical is 60sin 45. Won't those stay the same in a stationary frame?

its telling me the answer for b is 22.4 m/s in the opposite direction of the submarine though

ok... i think i see what you're saying. the x component of velocity if it looked like it was going straight up would be 0? because its 60cos(90)?

Homework Statement 1.) A projectile is fired from a submarine traveling horizontally at 20 m/s with respect to the water as shown in the figure below. According to an observer on the submarine, the projectile is fired at 45° with an initial velocity of 60 m/s. After firing the projectile, the...

ok I see. thanks for the help

switch the x and 2y so you have y=x+2 and -2y=-x-2. the x and 2 cancel so you have y=0 plug that back into find x. I still do not understand how I could use a system for this problem though.

yes I understand but for fy i have Nf=mgcos30-ftsin30 but since I don't know the force of tension I can't find Nf and in the x direction forces are mgsin30+ftcos30=ffric. I need the force of friction here to find ft but I need Nf for that which is why I am stuck.

wait I just realized I have 3 unknowns. Nf, Ftsin30 and Ftcos30. I have been trying to solve for them but can't seem to find a way too. I'm racking my brain trying to figure out how to find Normal force specifically but i don't see a way to manipulate it since we don't know ftsin30

yes in my work I had +ftcos30 but I just realized that I could easily solve for it now! thanks a lot!

so i started working it out and I realized there is still a problem! first i did sum of f in x direction = 0 and i got mgsin30-Ffriction-Ftcos30=0 the sum in the y direction is Nf-mgcos30+ftsin30=0. Now I believe I need solve for ftsin30 and ftcos30 and then add them as vectors but in order to...

my gameplan after finding out what the angle was looked like this. the normal force becomes mgcos30 for the block. then I can multiply us to the normal force to find the force of static friction needed to keep the box up. and then I go over to the sum of in the x direction and use that friction...

the question is What is Tmax the maximum value of T for which the block can be held in place with static friction? [FONT=Times]

ok so I think that because the axis is tilted 30 degrees down the component along the incline's angle would have to be ftcos30 because the tension is completely horizontal. the component perpendicular to this angle would be ftsin 30. I'm pretty unsure of myself on this concept though

ok I think I see your point. So it would also be a 30 degree angle between tension and the x-axis correct? The way I am thinking about it is if we made the angle of incline 0 what would the tension be. and I'm thinking it would be either 30 or 60 but i'd like a way in which I can tell for sure