Four forces act on an object, given by a = 33.3 n east, b = 30.3 n north, c = 70.7 n west, and d = 92.7 n south. (assume east and north are directed along the +x-axis and +y-axis, respectively.) hint (a) what is the magnitude of the net force (in n) on the object? n (b) what is the direction of the force? (enter your answer in degrees counterclockwise from the +x-axis.) ° counterclockwise from the +x-axis

a)   F = 72.75 N  and b)   T = 239º

Explanation:

In that exercise they ask us for the resultant force, the forces are vectors, for which their components are due. Let's start by writing the four forces in the xy chain system

       A = 33.3N i ^

       B = 30.3 N j ^

       C = -70.7 N i ^

       D = -92.7N j ^

Let's make the sum of each component

      Fₓ = A + C

      Fₓ = 33.3 -70.7

      Fₓ = -37.4 N

      = B + D

      = 30.3 -92.7

      = -62.4 N

Let's use the Pythagorean theorem and trigonometry to find the magnitude and angle

    F² = Fx² + Fy²

    F =√ [(-37.4² + (-62.4)² ] = √ 5292.52

    F = 72.75 N

    tan θ₁ = / Fₓ

    tan θ₁ = -62.4 / (- 37.42) = 1.6676

    θ₁ = 59º

Since the magnitudes of the two components of the force are negative, the angle is in the third quadrant, so to measure the x-axis, we must add 180º to the angle

     T = 180 + 59

    T = 239º

f=ma newton's second law 7x8=56n