Recent content by deserthobo

oooooh thanks a ton denverdoc, i guess i needed to read the problem more carefully. it was just confusing as rather the cart she was pushing was 29 degrees below the horizontal or the force she exerts on the cart is 29 degrees below...

no picture to go with the problem.

it says that the woman is pushing the cart in a direction 29 degrees below the horizontal

i really have no clue as to the direction of the frictional force are the two other forces I am missing the normal force and w=mg? i don't i need those in this problem.

hey guys i drew a free body diagram, can you tell me what's wrong http://img88.imageshack.us/img88/2869/freebodyhu4.jpg

wait i don't get it, doesn't the force of friction act opposite of the direction of the displacement and force that the person exerts? if that's the case then why am i using this ratio?

i don't know a! my guess is a = 48 N but my textbook says 54.9. how the heck did it get that answer?

Homework Statement http://img406.imageshack.us/img406/117/physicsprob3pp6.jpg known date 1. angle= 29 degrees 2. frictional force = 48 N 3. mass of cart = 16 kg 4. displacement = 22 meters 5. velocity = constant Homework Equations F=ma F(s)=mas W=F cos theta (s) The Attempt...

ok so i did everything following what mybsaccownt said and in the end i came up with 43 degrees. the book says 42 so this method is incorrect!

ok guys, if i plug in the horizontal velocity component which is 208 for vf and vo for x=1/2(vf+vo)t i will get 2704 m, which when plugged into arc tan (y/x) will give me the angle of 41..5 which when rounded up is 42 degrees. but does it make sense to plug in 208 for both vf and vo?

ok using vf=vo + at i got 12 s i plugged 12 s into x = 1/2(vf + vo)t and then found the arc tan with that number and came up with 62 degrees. doesn't work out.

hmm why do i want the arc tan of the final velocities and not the arc tan of the distances? also, arent the final velocites both = 0 since the package hits the ground

Homework Statement http://img413.imageshack.us/img413/2919/physicsprobpr6.jpg Xy=2400 m Homework Equations Xy=Voyt + 1/2at^2 Xx=1/2(Vox+Vx)t The Attempt at a Solution I calculated the vertical and horizontal velocity components of the given resultant velocity of 240 m/s...