Question

Three forces act on an object (at the origin of a rectangular coordinate system). Force one, F1, has a magnitude of 5.81 N and a direction Theta1 = 77.0 degrees, force two, F2, has a magnitude of 4.88 N and a direction of Theta2, = 156 degrees, and a force three F3, has a magnitude of 4.52 N and a direction of Theta 3 = 289 degrees. add these three vectors using the component method of vector addition. Call the resultant vector F and write the answer in component form (ie using unit vectors). then calculate the magnitude and direction of the vector F.

Answer 1

Answer:

F = -1.682 i + 3.374 j [N] ; F = 3.77[N]; α = 65.78°

Explanation:

In order to be able to understand this problem more easily, we will have to draw the vectors at the point of origin and in this way, with the help of the angles we will be able to find each of the components of the forces.

Looking at the attached image we can see each of the three forces, take each force and decompose them into the x & y axes.

F1 = 5.81 [N]

$F_{1x} = 5.81*cos(77) = 1.306[N]\\F_{1y} = 5.81*sin(77) = 5.66[N]$

F2= 4.88 [N]

$F_{2x}= - 4.88*cos(180-156) =-4.458[N]\\F_{2y}= 4.88*sin(180-156) =1.984[N]$

F3= 4.52 [N]

$F_{3x} = 4.52*cos(360-289)=1.47[N]\\F_{3y} = - 4.52*sin(360-289) = -4.27[N]$

Now we can sum each of the forces in the different components

$F_{x} =F_{x1}+F_{x2}+F_{x3} = 1.306-4.458+1.47 = - 1.682[N]\\F_{y} =F_{y1}+F_{y2}+F_{y3} = 5.66+1.984-4.27 = 3.374[N]$

F = -1.682 i + 3.374 j [N]

The total magnitude can be calculated by Pythagoras theorem

$F = \sqrt{(1.682)^{2}+(3.374)^{2} } \\F = 3.77 [N]$

The direction can be calculates as follows:

tan(α) = 3.374/1.682

α = 65.78°

This angle was calculated with respect to the horizontal