Question

the diagram shown below is shown for an object in equilibrium. write an equilibrium equation for the x direction (left = right) for the forces shown below.
image of forces f₁, f₂, f₃ with angles θ and φ
options:

• $f_1 \cos(\theta) + f_2 \cos(\phi) = f_3$
• $f_1 \cos(\theta) = f_2 \cos(\phi)$
• $f_1 \sin(\theta) = f_2 \sin(\phi)$
• $f_1 \sin(\theta) + f_2 \sin(\phi) = f_3$

Explanation:

Step1: Analyze Force Components in x-direction

For an object in equilibrium, the sum of forces in the x - direction (left - right) should be zero. We need to find the x - components of \(F_1\), \(F_2\) and \(F_3\) (but \(F_3\) is along the y - direction, so its x - component is 0). The x - component of \(F_1\) is \(F_1\sin(\theta)\) (assuming the angle \(\theta\) is measured from the y - axis? Wait, no, wait. Wait, if we consider the x - direction (horizontal), then for \(F_1\), the horizontal (x) component: if \(\theta\) is the angle with the y - axis, then the x - component is \(F_1\sin(\theta)\) (to the right), and for \(F_2\), the x - component is \(F_2\sin(\phi)\) (to the left). In equilibrium, the sum of forces to the right equals the sum of forces to the left. So \(F_1\sin(\theta)=F_2\sin(\phi)\)? Wait, no, wait, maybe I got the angle wrong. Wait, maybe \(\theta\) is the angle with the x - axis? No, the diagram shows \(F_1\) and \(F_2\) making angles \(\theta\) and \(\phi\) with the vertical (since \(F_3\) is vertical). So the horizontal (x) component of \(F_1\) is \(F_1\sin(\theta)\) (right - ward) and the horizontal component of \(F_2\) is \(F_2\sin(\phi)\) (left - ward). For equilibrium in x - direction, the net force is zero, so \(F_1\sin(\theta)-F_2\sin(\phi) = 0\), which implies \(F_1\sin(\theta)=F_2\sin(\phi)\). Wait, but let's check the options. The third option is \(F_1\sin(\theta)=F_2\sin(\phi)\), the second option is \(F_1\cos(\theta)=F_2\cos(\phi)\) (which would be for y - direction if angles are with x - axis, but here angles are with y - axis). Wait, maybe I misassigned the angle. Let's re - consider: if \(\theta\) is the angle between \(F_1\) and the x - axis, then the x - component of \(F_1\) is \(F_1\cos(\theta)\) and y - component is \(F_1\sin(\theta)\). Similarly, for \(F_2\), if \(\phi\) is the angle with the x - axis (on the left), then x - component is \(F_2\cos(\phi)\) (but in the opposite direction). But in the diagram, \(F_3\) is vertical, so \(F_1\) and \(F_2\) are in the upper half, making angles with the vertical. So the horizontal (x) components: \(F_1\) has a horizontal component of \(F_1\sin(\theta)\) (right), \(F_2\) has a horizontal component of \(F_2\sin(\phi)\) (left). For equilibrium, right forces = left forces, so \(F_1\sin(\theta)=F_2\sin(\phi)\). Wait, but let's check the options. The third option is \(F_1\sin(\theta)=F_2\sin(\phi)\), the second is \(F_1\cos(\theta)=F_2\cos(\phi)\) (which would be for y - direction: \(F_1\cos(\theta)+F_2\cos(\phi)=F_3\) for y - direction). Wait, the question is about x - direction. So in x - direction, the forces from \(F_1\) (right) and \(F_2\) (left) must balance. So \(F_1\sin(\theta)=F_2\sin(\phi)\) (since \(F_3\) has no x - component). Let's check the options:

• Option 1: \(F_1\cos(\theta)+F_2\cos(\phi)=F_3\) → this is y - direction equilibrium.
• Option 2: \(F_1\cos(\theta)=F_2\cos(\phi)\) → y - direction (if angles are with x - axis, but here angles are with y - axis, so no).
• Option 3: \(F_1\sin(\theta)=F_2\sin(\phi)\) → x - direction equilibrium (right component of \(F_1\) equals left component of \(F_2\)).
• Option 4: \(F_1\sin(\theta)+F_2\sin(\phi)=F_3\) → wrong, \(F_3\) is y - direction, and x - components can't sum to y - component.

So the correct equation for x - direction equilibrium is \(F_1\sin(\theta)=F_2\sin(\phi)\).

Step2: Match with Options

Looking at the options, the third option is \(F_1\sin(\theta)=F_2\sin(\phi)\).

Answer:

\(F_1\sin(\theta)=F_2\sin(\phi)\) (the third option: \(F_1\sin(\theta)=F_2\sin(\phi)\))