Question

a force vector points at 25 degrees up from the horizontal. the y component of the force vector is equal to 1 newton. what must be the magnitude of the full force vector?
1 newton
1.89 newtons
1.18 newtons
2.18 newtons
2.37 newtons

Explanation:

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

The y - component of a vector \(\vec{F}\) with magnitude \(F\) and angle \(\theta\) (measured from the horizontal) is given by \(F_y = F\sin\theta\). We know that \(F_y = 1\space N\) and \(\theta=25^{\circ}\), and we need to find \(F\).

Step2: Rearrange the formula to solve for \(F\)

From \(F_y = F\sin\theta\), we can solve for \(F\) by dividing both sides of the equation by \(\sin\theta\). So \(F=\frac{F_y}{\sin\theta}\).

Step3: Substitute the known values

We know that \(F_y = 1\space N\) and \(\theta = 25^{\circ}\). \(\sin(25^{\circ})\approx0.4226\). Then \(F=\frac{1}{0.4226}\approx2.37\space N\)? Wait, no, wait. Wait, let's recalculate. Wait, \(\sin(25^{\circ})\) is approximately \(0.4226\)? Wait, no, wait, \(\sin(25^{\circ})\approx0.4226\)? Wait, no, actually, \(\sin(25^{\circ})\approx0.4226\)? Wait, no, let's use a calculator. \(\sin(25^{\circ})\): 25 degrees, sine of 25 is approximately 0.4226. Wait, but if \(F_y = F\sin\theta\), then \(F=\frac{F_y}{\sin\theta}\). So if \(F_y = 1\), then \(F=\frac{1}{\sin(25^{\circ})}\approx\frac{1}{0.4226}\approx2.37\)? Wait, but let's check the options. Wait, maybe I made a mistake. Wait, no, wait, the angle is 25 degrees from the horizontal, so the y - component is \(F\sin\theta\). So \(F=\frac{F_y}{\sin\theta}\). Let's compute \(\sin(25^{\circ})\) more accurately. Using a calculator, \(\sin(25^{\circ})\approx0.4226182617\). Then \(F = \frac{1}{0.4226182617}\approx2.366\approx2.37\space N\). Wait, but let's check the options. The last option is 2.37 Newtons. Wait, but let's re - check. Wait, maybe the angle is with respect to the vertical? No, the problem says 25 degrees up from the horizontal. So the y - component is \(F\sin\theta\). So the calculation is correct.

Answer:

2.37 Newtons (the option with "2.37 Newtons")