Question

the forces below act on an object that is in equilibrium in the y direction. f₂ has a magnitude of 33 and is at an angle of 57 degrees from the x axis. f₁ has a magnitude of 11 and is at an angle of 23 degrees from the x axis. calculate the magnitude of f₃ 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 \( F_1 \), \( F_1 = 11 \) and \( \theta = 23^\circ \). So, \( F_{1y}=11\times\sin(23^\circ) \). Calculating \( \sin(23^\circ)\approx0.3907 \), then \( F_{1y}\approx11\times0.3907 = 4.2977 \).

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

For \( F_2 \), the angle from the x - axis is \( 57^\circ \), but since it is on the other side of the y - axis (from the diagram), the angle with respect to the positive y - direction? Wait, no, the formula for the y - component is still \( F_y = F\sin\theta \) where \( \theta \) is the angle from the x - axis. Wait, actually, looking at the diagram, \( F_2 \) is at an angle \( \phi \) from the negative x - axis? Wait, no, the problem says \( F_2 \) is at an angle of 57 degrees from the x - axis. So the y - component of \( F_2 \) is \( F_{2y}=33\times\sin(57^\circ) \). \( \sin(57^\circ)\approx0.8387 \), so \( F_{2y}\approx33\times0.8387 = 27.6771 \).

Step3: Apply equilibrium in y - direction

In equilibrium, the sum of the forces in the y - direction is zero. Let's assume the positive y - direction is upward. Then \( F_{1y}+F_{2y}-F_3 = 0 \) (because \( F_3 \) is downward, so its y - component is negative if we take upward as positive). So \( F_3=F_{1y}+F_{2y} \).

Step4: Calculate \( F_3 \)

Substitute the values of \( F_{1y} \) and \( F_{2y} \) into the equation: \( F_3=4.2977 + 27.6771=31.9748\approx32.0 \) (rounded to a reasonable decimal place).

Answer:

The magnitude of \( F_3 \) is approximately \( 32.0 \) (the more precise value can be calculated as \( 11\sin(23^\circ)+33\sin(57^\circ)\approx11\times0.3907 + 33\times0.8387=4.2977+27.6771 = 31.9748\approx32 \)).