Asked by Derek
A force of magnitude 15 N is the resultant of two forces, one of which has a magnitude of 8 N and acts at an angle of 30° to the resultant. Find the magnitude and direction of the other force...how do I begin to solve this!?
Answers
Answered by
bobpursley
Law of cosines:
The side you are looking for is opposite the 30 degree angle.
c^2=15^2+8^2-2*15*8cos30
I get a magnitude of about 9
For the angle, Iwould use law of sines. Draw a sketch first.
The side you are looking for is opposite the 30 degree angle.
c^2=15^2+8^2-2*15*8cos30
I get a magnitude of about 9
For the angle, Iwould use law of sines. Draw a sketch first.
Answered by
Damon
Well, I am taking the resultant to be along the x axis and the 8 N vector at 30 degrees above the x axis.
Then our vector of length x will lie below the x axis an angle y
Now there are two things that have to be true
8 cos 30 + x cos y = 15
8 sin 30 - x sin y = 0
or
6.93 + x cos y = 15
4 - x sin y = 0
or
x cos y = 8.07
x sin y = 4
but we know sin y / cos y = tan y so divide
tan y 4/8.07
y = 26.4 degrees below x axis
which is
26.4 + 30 = 56.4 degrees from original vector
x = 4/sin26.4 = 9 Newtons
Then our vector of length x will lie below the x axis an angle y
Now there are two things that have to be true
8 cos 30 + x cos y = 15
8 sin 30 - x sin y = 0
or
6.93 + x cos y = 15
4 - x sin y = 0
or
x cos y = 8.07
x sin y = 4
but we know sin y / cos y = tan y so divide
tan y 4/8.07
y = 26.4 degrees below x axis
which is
26.4 + 30 = 56.4 degrees from original vector
x = 4/sin26.4 = 9 Newtons
Answered by
Count Iblis
You can also solve this problem purely algebraically without using geometry or drawing diagrams. If we denote the resultant force by R, the force of magnitude 8 N at an angle of 30° to the resultant by y and the unknown force by x, we have:
x + y = R ------->
x = R - y
Square both sides:
x^2 = R^2 + y^2 - 2 R dot y
The inner product R dot y can be expressed as:
R dot y = |R||y| cos(30°)
This then gives |x| = about 9 N
To find the angle of x to the resultant (let's call this theta), you can take the inner product of x with R and use that
x dot R = |x| |R| cos(theta)
We have:
x = R - y
taking the inner product of both sides with R gives:
x dot R = R^2 - y dot R =
R^2 -|y||R|cos(30°)
If we divide both sides by |x||R| we get cos(theta) on the l.h.s.:
cos(theta) = [|R| - |y|cos(30°)]/|x|
this gives theta = 26.4°
x + y = R ------->
x = R - y
Square both sides:
x^2 = R^2 + y^2 - 2 R dot y
The inner product R dot y can be expressed as:
R dot y = |R||y| cos(30°)
This then gives |x| = about 9 N
To find the angle of x to the resultant (let's call this theta), you can take the inner product of x with R and use that
x dot R = |x| |R| cos(theta)
We have:
x = R - y
taking the inner product of both sides with R gives:
x dot R = R^2 - y dot R =
R^2 -|y||R|cos(30°)
If we divide both sides by |x||R| we get cos(theta) on the l.h.s.:
cos(theta) = [|R| - |y|cos(30°)]/|x|
this gives theta = 26.4°
Answered by
Derek
My diagram doesn't work out, I have done it to scale, but I can't figure out how to draw it. I got the same answers as everyone above, I used the cosine and sine law to do mine (I got my answer after I posted :) ). But my diagram looks weird. Any thoughts?
Answered by
Tapiwa Mudzori
15 cos
Answered by
Anonymous
two forces of magnitude 10N and 30N act upon a body in direction making angle of 30 and 45 with x axis find the ressultant force?
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