The angle between two vectors (F₁ and F₂) is 90-28=62°
The second angle of the force parallelogram is (360 -2•62)/2= 118°.
Cosine law gives
F=sqrt (F₁² +F₂² -F₁•F₂•cos118°)= 9.2
tan α = (5+5.9•sin28°)/5.9•cos28° =1.49
α = 56.2°
A force F 1 of magnitude 5.90 units acts on an object at the origin in a direction θ = 28.0° above the positive x-axis. (See the figure below.) A second force F 2 of magnitude 5.00 units acts on the object in the direction of the positive y-axis. Find graphically the magnitude and direction of the resultant force F 1 + F 2.
magnitude___________? units
direction __________? counterclockwise from the +x-axis
(I tried doing 5.9xcos(28)=5.21N and 5.9xcos(28)+5=10.2N. When I divided them the answers that I got were 11.45 and 63 degrees and they were incorrect I think that I set the problem up incorrectly.
2 answers
Fr = 5.9[28o] + 5[90o].
X = 5.9*Cos28 = 5.2Units.
Y = 5*sin28 + 5*sin90 = 7.35 Units.
Fr = sqrt(x^2+y^2) =
Tan A = Y/X = 7.35/5.2 = 1.41346,
A = 54.7o = The direction.
X = 5.9*Cos28 = 5.2Units.
Y = 5*sin28 + 5*sin90 = 7.35 Units.
Fr = sqrt(x^2+y^2) =
Tan A = Y/X = 7.35/5.2 = 1.41346,
A = 54.7o = The direction.