Physics 11 Help: Solving a Three-Force Problem for Point J

[Iceclover]

i was just wondering if anyone could help me work out this probelm? I've been having an extremely difficult time in physics. My teacher has only taught it once before and is difficult to understand. Anyways, here's the probelm. Three forces act simultaneously on point J. One force is 10.0 N north; the second is 15.0 N west; the third is 15.0N 30.0 degrees east of north. Determine the magnitude and direction of the resultant force.

 

[PhanthomJay]

i was just wondering if anyone could help me work out this probelm? I've been having an extremely difficult time in physics. My teacher has only taught it once before and is difficult to understand. Anyways, here's the probelm. Three forces act simultaneously on point J. One force is 10.0 N north; the second is 15.0 N west; the third is 15.0N 30.0 degrees east of north. Determine the magnitude and direction of the resultant force.

The first step is to find the x and y components of each vector, paying heed as to whether they are plus or minus, then add all the x components together, and , separately, add all the y components together. Can you get that far? I'll start you off: the first vector is straight up north, in the y direction, so it has no x component. Thus, it is 0 Newtons in the x direction, and +10 N in the y direction.

 

[Iceclover]

thank you, so i came up with 24.2N but now how do i find the direction in degrees?

 

[HalfManHalfAmazing]

You have to use the that good ol' sin, cos, tan stuff to find it out. Like if you have the x and y components you have a right angle triangle - this is used to find the angle. this link should help a bit (ignore the graphs of the functions for now) http://www.gcseguide.co.uk/sin,_cos,_tan.htm

 

[Iceclover]

yeah i used sin cos and tan. This is what i did. y=sin(30degrees)(10)=5N
x=cos(30degrees)(10)=8.6N
8.6+15=23.66

a^2+b^2=c^2
23.66^25^2=24.4

now how do i find the direction?

 

[Max Eilerson]

You've got your cos and sin mixed up.

y-component = 10N + 15cos (30) =23 N (North)
x-component = 15N - 15sin(30) = 7.5 N (West)
remember West and East are opposite directions hence the minus.

magnitude is sqrt(23^2 + 7.5^2) = 24.2 N

To find the angle you need to use either the cos or sin relation and solve for theta.

 

[Iceclover]

where do you find the other 15 to add to the 10 in 10+15cos(30)

 

[Max Eilerson]

'15.0N 30.0 degrees east of north'

This vector has a y-component of 15cos (30) (North), and an x-component of 15sin(30) (East). To check this is right

sqrt[(15cos 30)^2 + (15sin 30)^2] = 15 N, which is the magnitude of original vector I just resolved.

 

[Iceclover]

Ok so i gotall that and i drew a picture. I managed to get the direction earlier and it was 108degrees but now i can't remember how i got that answer.

 

[Max Eilerson]

108degress is correct if you say 108degrees North of East. Here though your final vectors and pointing North and West, so it would be better too say 18degrees West of North.

[tex] /theta = tan^{-1} (y-component)/(x-component) = tan^{-1} (23 N / 7.5 N) = 72degrees. [/tex] That gives you the answer North of West, then 90 -72 =18 degrees, for West of North.