An object in equilibrium has three forces exerted on it. A 36 N force acts at 90° from the x axis and a 42 N force acts at 60°. What are the magnitude and direction of the third force?
To solve this problem, we can use the concept of vector addition and equilibrium conditions. In equilibrium, the net force acting on an object is zero.
Since we have two forces, we can represent them as vectors using their magnitudes and directions. Let's call the first force F1 and the second force F2.
1. F1 = 36 N at 90° from the x-axis. This force can be represented as F1 = 36 N (cos 90° î + sin 90° ȷ̂).
2. F2 = 42 N at 60°. This force can be represented as F2 = 42 N (cos 60° î + sin 60° ȷ̂).
Now, we need to find the magnitude and direction of the third force F3.
Since the object is in equilibrium, the net force acting on it is zero. Therefore, F1 + F2 + F3 = 0.
To find F3, we rearrange the equation:
F3 = -F1 - F2
Substituting the values:
F3 = -(36 N (cos 90° î + sin 90° ȷ̂)) - (42 N (cos 60° î + sin 60° ȷ̂))
Now, we can calculate F3 by evaluating the expression.
F3 = -(-36 N (î + ȷ̂) - 42 N (cos 60° î + sin 60° ȷ̂))
F3 = 36 N (î + ȷ̂) + 42 N (cos 60° î + sin 60° ȷ̂)
F3 = 36 N î + 36 N ȷ̂ + 42 N (cos 60° î + sin 60° ȷ̂)
Simplifying further, we get:
F3 = (36 + 42 cos 60°) N î + (36 + 42 sin 60°) N ȷ̂
Now, we can find the magnitude and direction of F3.
Magnitude of F3 = sqrt[(36 + 42 cos 60°)^2 + (36 + 42 sin 60°)^2]
Direction of F3 = arctan[(36 + 42 sin 60°) / (36 + 42 cos 60°)]
Evaluating these expressions will give us the magnitude and direction of the third force.
ΣF(x) =F1(x) +F2(x) +F3(x) =0
0+42•cos60°+F3(x) =0
F3(x) = -42•cos60°= -21 N
ΣF(y) =F1(y) +F2(y) +F3(y) =0
36+42•sin60°+F3(y) =0
F3(y) =-36 -42•sin60°= -72.4 N
F=sqrt{F(x)²+F(y)²}
tan θ=F(y)/F(x)