A force of 200n is inclined at an angle of 120° to another force P. The angle between the 200n force and the resultant force is 50°. Find the magnitude of the force P and the resultant of the two force.?

Answers:  

Magnitude of force P = 163.041494

Magnitude of resultant force = 184.320997

Values are approximate. Units are in newtons.

Explanation:

Let for P be pulled directly to the east. The vector for this force is <x, 0> where x is positive.

The 200 newton force has the vector <200*cos(120), 200*sin(120)>

The resultant vector is <x+200*cos(120),200*sin(120)>. Each component is the sum of the corresponding components of <x,0> and <200*cos(120), 200*sin(120)>

The resultant vector is also <r*cos(70),r*sin(70)>. Note how 70+50 = 120. The 50 degree angle is known, so we effectively do 120-50 = 70 to find the angle of the resultant vector with the positive x axis.

The resultant vector expressions we found were

Equate the y components of each resultant vector expression. Solve for r

r*sin(70) = 200*sin(120)

r = 200*sin(120)/sin(70)

r = 184.320997021376

Make sure your calculator is in degree mode.

Let's round this r value to 6 decimal places to simplify things a bit

r = 184.320997

Now equate the x components of each resultant vector expression, plug in the r value we found, and solve for x

x+200*cos(120) = r*cos(70)

x+200*cos(120) = 184.320997*cos(70)

x = 184.320997*cos(70) - 200*cos(120)

x = 163.041494

this value is approximate just like r is as well

The magnitude of force P is

magnitude = sqrt(x^2+y^2)

magnitude = sqrt(x^2+0^2)

magnitude = sqrt((163.041494)^2+0^2)

magnitude = 163.041494

Which is equal to the x value. This applies because y = 0.

The magnitude of the resultant is r = 184.320997 which we found earlier

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