Show that the x-component of the tension force Fc in terms of its magnitude Fc is given by -0.3420 Fc
To show that the x-component of the tension force Fc in terms of its magnitude Fc is given by -0.3420 Fc, we use vector operations and trigonometry.
Let's consider a coordinate system where the x-axis is horizontal and the y-axis is vertical. The tension force Fc can be represented as a vector with two components: Fc_x (the x-component) and Fc_y (the y-component).
We can express the magnitude of the tension force Fc as:
|Fc| = √(Fc_x^2 + Fc_y^2)
Since we want to express the x-component of the tension force Fc in terms of its magnitude Fc, we need to find Fc_x in terms of Fc. To do this, we can use trigonometry.
Let's assume that the angle between the tension force Fc and the positive x-axis is θ. We can then write:
Fc_x = Fc * cos(θ)
Given that Fc_x is equal to -0.3420 Fc, we can set up the equation as:
-0.3420 Fc = Fc * cos(θ)
Now, we can solve this equation to find the value of cos(θ) and determine the relationship between Fc_x and Fc.
Dividing both sides of the equation by Fc, we get:
-0.3420 = cos(θ)
To find the angle θ, we take the inverse cosine of both sides:
θ = cos^(-1)(-0.3420)
Evaluating this inverse cosine using a calculator, we find that θ is approximately 109.2 degrees.
Therefore, the x-component of the tension force Fc in terms of its magnitude Fc is given by:
Fc_x = Fc * cos(109.2°) = -0.3420 Fc