Finding a force...?

Question

I just cant find the right method to get this...

If youre told that 2 forces will give a resultant force and you know the magnitude of only one of the 2 forces, the angle between that force and the resulting force, and the magnitude of the resulting force how do you find the other force????
Heres the exact question:
the force P has magnitude 5N and acts due north. The resultant P+Q has manitude 9N and acts in the direction 060 degrees.Calculate the magnitude and direction of Q...

BDW, The resultant force is not equal to P + Q

answers should be: sqrt 61 Neutons...and 093.7 degrees (ie from North)

Help! Im studying and just cant get this one

Answer 1

Make a triangle of forces: AB is P force from A to B with amplitude 5 N heading north. AC is the resultant force with magnitude 9 N, 60 o from the North. Then BC is Q. Now you have a triangle ABC where AB and AC and angle A are known. Then apply cos theorem:
BC^2 = AB^2 + AC^2 - AB*AC*cos(A), and you get
BC^2 = 25 + 81 - 45 = 61.
Therefore magnitude of Q = sqrt(61).
The Q angle you can calculate from:
|Q|*cos(alpha) = |P+Q|*cos(30) => alpha = acos(9*sqrt(3)/(2*sqrt(61))) = 3.7 degrees (from the East to North).

Answer 2

P + Q = T; where each term is a force vector.

Convert all the vectors to X,Y components to put them on the same coordinates before trying to solve. For example:

Px = P cos(theta) and Py = P sin(theta); theta is relative to the X axis; theta = 90 deg (due north) and magnitude |P| = 5 N. Thus, by inspection, Py = |P| = 5 N and Px = 0.

Qx = Q cos(omega) and Qy = Q sin(omega); omega is relative to the same X axis as theta is.

The vector T = Px + Qx + Py + Qy; so its magnitude is |T| = sqrt[(Px + Qx)^2 + (Py + Qy)^2] at an angle psi = 60 deg = arcsin((P + Qy)/T); where |T| = 9 N, |P| = 5 N, and you can find Qy from this. Then, using |T| = sqrt[(Px + Qx)^2 + (Py + Qy)^2] = sqrt((Qx)^2 + (P + Qy)^2); you can solve for Qx. Thus, you have all X and Y components of P and Q.

Thus, omega = arcsin[Qy/sqrt(Qx^2 + Qy^2)] and |Q| = sqrt(Qx^2 + Qy^2).

Note, I used standard cartesian coordinates with zero angle along the X axis on the positive side; so north = 90 deg, west = 180 deg, etc. To get angles with north = 0, just subtract 90 degs from the answers you get from the above relationships.

BDW: the resultant force T is equal to the sum of the two vectors P + Q, that was a given.

Answer 3

You know that
R = P + Q
Q = R - P

You convert forces into rectangular coordinates:

P = |P|cos(p)X + |P|sin(p)Y
Q = |Q|cos(q)X + |Q|sin(q)Y
R = |R|cos(r)X + |R|sin(r)Y

Now

|Q|cos(q) = |R|cos(r) - |P|cos(p)
|Q|sin(q) = |R|sin(r) - |P|sin(p)

Once you get this you can convert back to polar coordinates

|Q| = sqrt((|Q|cos(q))^2 + (|Q|sin(q))^2)
and q = invcos(|Q|cos(q)/|Q|)