Question

the forces below act on an object that is in equilibrium in the y direction. $f_2$ has a magnitude of 37 and is at an angle of 55 degrees from the x axis. $f_1$ has a magnitude of 14 and is at an angle of 23 degrees from the x axis. calculate the magnitude of $f_3$ needed to keep the object in equilibrium in the y direction. note: the object may not be in equilibrium in the x direction, but you dont need to solve for any of the x components for this problem.

Explanation:

Step1: Find y - component of \( F_1 \)

The formula for the y - component of a force \( F \) with magnitude \( F \) and angle \( \theta \) from the x - axis is \( F_y=F\sin\theta \). For \( F_1 \), \( F_1 = 14 \) and \( \theta=23^{\circ} \). So, \( F_{1y}=14\sin(23^{\circ}) \).
Calculating \( \sin(23^{\circ})\approx0.3907 \), then \( F_{1y}=14\times0.3907\approx5.47 \).

Step2: Find y - component of \( F_2 \)

For \( F_2 \), magnitude \( F_2 = 37 \) and angle from x - axis is \( 55^{\circ} \). Using the formula \( F_y = F\sin\theta \), we have \( F_{2y}=37\sin(55^{\circ}) \).
Calculating \( \sin(55^{\circ})\approx0.8192 \), then \( F_{2y}=37\times0.8192\approx30.31 \).

Step3: Apply equilibrium condition in y - direction

In equilibrium in the y - direction, the sum of the upward forces is equal to the sum of the downward forces. Let the upward direction be positive and downward be negative. The forces \( F_{1y} \) and \( F_{2y} \) are upward, and \( F_3 \) is downward. So, \( F_{1y}+F_{2y}-F_3 = 0 \) (since net force in y - direction is zero).
Rearranging for \( F_3 \), we get \( F_3=F_{1y}+F_{2y} \).
Substituting the values of \( F_{1y} \) and \( F_{2y} \): \( F_3=5.47 + 30.31=35.78\approx35.8 \) (or more accurately, using more precise values of sine functions)

Let's recalculate with more precise values:
\( \sin(23^{\circ})=\sin(23)\approx0.3907311285 \)
\( F_{1y}=14\times0.3907311285 = 5.470235799 \)
\( \sin(55^{\circ})=\sin(55)\approx0.8191520443 \)
\( F_{2y}=37\times0.8191520443=30.30862564 \)
\( F_3=F_{1y}+F_{2y}=5.470235799 + 30.30862564=35.77886144\approx35.8 \) (or if we consider significant figures, maybe 36, but let's check the calculation again)

Wait, actually, the angle for \( F_2 \): looking at the diagram, the angle \( \phi \) is the angle between \( F_2 \) and the negative y - axis? Wait, no, the problem says "at an angle of 55 degrees from the x axis". Wait, maybe I made a mistake in the angle interpretation. Wait, if the force \( F_2 \) is on the left side, the angle with the y - axis? Wait, no, the problem says "at an angle of 55 degrees from the x axis". So if the force is on the left, the angle between \( F_2 \) and the x - axis is \( 55^{\circ} \), so the angle with the y - axis is \( 90 - 55=35^{\circ} \)? Wait, no, no. The formula \( F_y=F\sin\theta \) where \( \theta \) is the angle with the x - axis. So if the force is above the x - axis, the y - component is \( F\sin\theta \). But in the diagram, \( F_2 \) is on the left, so maybe the angle with the negative x - axis? Wait, no, the problem says "at an angle of 55 degrees from the x axis". So regardless of left or right, the angle from the positive x - axis. So if \( F_2 \) is on the left, its angle from the positive x - axis is \( 180 - 55 = 125^{\circ} \)? Wait, that's a mistake! Oh no, I misinterpreted the angle. If \( F_2 \) is on the left side of the y - axis, the angle from the positive x - axis is \( 90 + (90 - 55)=125^{\circ} \)? Wait, no. Let's look at the diagram: \( F_1 \) is on the right, angle \( \theta \) with positive x - axis, \( F_2 \) is on the left, angle \( \phi \) with negative x - axis? Wait, the problem says " \( F_2 \) has a magnitude of 37 and is at an angle of 55 degrees from the x axis". So maybe the angle between \( F_2 \) and the positive x - axis is \( 180 - 55=125^{\circ} \)? Wait, no, if the force is going up and to the left, the angle from the positive x - axis is \( 180 - \alpha \), where \( \alpha \) is the angle with the negative x - axis. But the y - component of a force with angle \( \theta \) from the positive x - axis is \( F\sin\theta \). If \( \theta = 180 - 55=125^{\circ} \), then \( \sin(125^{\circ})=\sin(55^{\circ})\approx0.8192 \), which is the same as before. Wait, because \( \sin(180 - x)=\sin(x) \). So my initial calculation of the y - component of \( F_2 \) was correct, because \( \sin(55^{\circ})=\sin(125^{\circ}) \). So the y - component of \( F_2 \) is still \( 37\sin(55^{\circ}) \) because the y - component depends on the sine of the angle with the x - axis, regardless of left or right (since sine is positive in both first and second quadrants).

So going back, the equilibrium condition: the sum of the y - components of \( F_1 \) and \( F_2 \) (upward) must equal the magnitude of \( F_3 \) (downward). So \( F_3=F_{1y}+F_{2y}=14\sin(23^{\circ})+37\sin(55^{\circ}) \)

Calculating this:

\( 14\sin(23^{\circ})=14\times0.3907311285 = 5.470235799 \)

\( 37\sin(55^{\circ})=37\times0.8191520443 = 30.30862564 \)

Adding them together: \( 5.470235799+30.30862564 = 35.77886144\approx35.8 \) (or 36 when rounded to a reasonable number of significant figures. The given values: \( F_1 = 14 \) (two significant figures?), \( F_2 = 37 \) (two significant figures?), angles are given as whole numbers. So maybe two significant figures? But 14 is two, 37 is two, angles are whole numbers. Wait, 14 could be two, 37 two. Then \( 14\sin(23^{\circ})\approx14\times0.39 = 5.46 \), \( 37\sin(55^{\circ})\approx37\times0.82 = 30.34 \), sum is \( 5.46 + 30.34=35.8\approx36 \). But let's check with more accurate calculation.

Using a calculator for \( 14\sin(23^{\circ})+37\sin(55^{\circ}) \):

\( \sin(23^{\circ})\approx0.3907 \), \( 14\times0.3907 = 5.47 \)

\( \sin(55^{\circ})\approx0.8192 \), \( 37\times0.8192 = 30.31 \)

\( 5.47+30.31 = 35.78\approx35.8 \)

Answer:

\( \approx35.8 \) (or more precisely, approximately 35.8, if we use more decimal places, it's approximately 35.78)