Question

a force vector shown below has a magnitude of 40 newtons. the force vector is at an angle of 65 degrees from the x axis. determine the magnitude of the y component of the force vector.

Explanation:

Step1: Recall the formula for the y - component of a vector

The formula for the y - component (\(F_y\)) of a vector \(\vec{F}\) with magnitude \(F\) and angle \(\theta\) (measured from the x - axis) is \(F_y=F\sin\theta\).

Step2: Substitute the given values

We are given that \(F = 40\space N\) and \(\theta=65^{\circ}\). So we substitute these values into the formula: \(F_y = 40\times\sin(65^{\circ})\).
We know that \(\sin(65^{\circ})\approx0.9063\). Then \(F_y=40\times0.9063 = 36.252\space N\).

Answer:

The magnitude of the y - component of the force vector is approximately \(36.25\space N\) (or more precisely \(40\sin(65^{\circ})\approx36.25\space N\))