Quick question about bearings in M1

A particle P, of mass 3 kg, moves under the action of two constant forces (6i + 2j) N and (3i - 5j) N.

a) Find in the form (ai + bj) N, the resultant force F acting on P.

b) Find, in degrees to one decimal place, the angle between F and j.

i got tan x = 9/3 for part b and then added 90 degrees. is that right? im not so sure about how u place the directions of the forces cuz this affects the angle u get.

guys plz help.

Reply 1

elpaw

F = 9i + 7j

|F| = root(81+49) = root(130)

cos@ = 7/root(130)

=> @ = arccos(7/root(130))
= 52.1 degrees

Reply 2

Col-C

stanny

A particle P, of mass 3 kg, moves under the action of two constant forces (6i + 2j) N and (3i + 5j) N.

a) Find in the form (ai + bj) N, the resultant force F acting on P.

b) Find, in degrees to one decimal place, the angle between F and j.

i got tan x = 9/3 for part b and then added 90 degrees. is that right? im not so sure about how u place the directions of the forces cuz this affects the angle u get.

guys plz help.

I think (I don't have a piece of paper to scribble on - this may be wrong)....

tan^-1 (9/7) is the angle between F and the x(i) axis

so 90 - tan^-1 is the angle between F and the y(j) axis

I get F to be (9i + 7j) but check your signs....if you meant to type -2i instead of +2i, then yes, it would be 90-tan^-1(9/3)

Reply 3

zoidberg

been a while since I did M1, but i think you just add them for the resultant force and then use pythagoras to find the angle between the vertical and the resultant force. tan@ = 9/7 ? i could be wrong

Reply 4

stanny

sorry there... its actually (3i - 5j) not (3i + 5j)... really sorry

Reply 5

Col-C

stanny

sorry there... its actually (3i - 5j) not (3i + 5j)... really sorry

You wrote 3i - 5j initially!

Am I going blind? Help me people. My brain has been over-calculusised

Reply 6

stanny

btw is it the angle between the j axis and the force?

the force is like south east... cuz (9i - 3j) is 9 across and 3 down...

so is it like the angle it makes with the j axis pointing upwards? so u get an angle more than 90.

so tan x = 9/3
x= 18.43...

so 90 +18.43... = 108.4 degrees (1dp) ??

Reply 7

elpaw

F = 9i - 3j

|F| = root(81+9) = 3root(10)

cos@ = -3/3root(10)

=> @ = arccos(-1/root(10))
= 108.4 degrees

Reply 8

stanny

Col-C

You wrote 3i - 5j initially!

Am I going blind? Help me people. My brain has been over-calculusised

hey ur not blind, im really sorry... cuz i jus found out from my paper and editted my msg...

Reply 9

Col-C

stanny

hey ur not blind, im really sorry... cuz i jus found out from my paper and editted my msg...

LOL - this explains why I appear dumb. Forgive me, digital electronics exam this morning. I can only count in 1's and zeroes :smile:

Let the others help you - I'm off for food :biggrin:

Reply 10

stanny

elpaw

F = 9i - 2j

|F| = root(81+4) = root(85)

cos@ = -2/root(85)

=> @ = arccos(-2/root(85))
= 102.5 degrees

isnt it (9i -3j)?

btw the real ans is 71.6... it actually joined a j vector to the end of the resultant force... is its like a triangle with an acute angle... argh this is confusing

Reply 11

elpaw

stanny

isnt it (9i -3j)?

btw the real ans is 71.6... it actually joined a j vector to the end of the resultant force... is its like a triangle with an acute angle... argh this is confusing