juan draws a free - body diagram of an object that is in dynamic equilibrium moving to the left. which labels correctly complete the diagram?
$f_f=-30n$
$f_g = y$
$f_t = x$
$f_n=45n$
$x:30n$
$y:-30n$
$x:30n$
$y:-45n$
$x:-45n$
$y:30n$
$x:45n$
$y:-45n$

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# Brief Explanations:
1. **Understand Force Equilibrium in Free - Body Diagrams**:
   - In a free - body diagram for an object in dynamic equilibrium (constant velocity, so net force is zero), the forces in the x - direction and y - direction must balance separately.
   - For the x - direction: The object is moving to the left. The friction force $F_f=- 30\ N$ (assuming left is negative or right is positive, but let's analyze the forces). The applied force in the x - direction ($F_t$) and the friction force should balance? Wait, no, looking at the diagram, the normal force $F_N = 45\ N$ is in the positive x - direction? Wait, no, maybe the coordinate system: Let's assume the upward is positive y and to the right is positive x. But the object is moving to the left, so the friction force $F_f=-30\ N$ (if left is negative). But for equilibrium, the sum of forces in x - direction: $F_N+F_f + F_t=0$? Wait, no, maybe the normal force is a typo? Wait, no, looking at the options, we need to find the correct $F_t$ (x - direction force) and $F_g$ (y - direction force, weight).
   - For the y - direction: The normal force (upward) and the weight (downward) should balance. The normal force $F_N = 45\ N$ is in the x - direction? Wait, no, maybe the labels are mixed. Wait, the diagram has $F_N = 45\ N$ (right - ward), $F_f=-30\ N$ (left - ward), $F_g$ (upward? No, weight is downward) and $F_t$ (downward? No, $F_t$ is in x - direction? Wait, maybe the diagram has:
     - Let's re - interpret the diagram: The vertical forces (y - direction): Let's say upward is positive y. The force $F_g$ is upward? No, weight is usually downward. Wait, maybe the normal force is upward (y - direction) and weight is downward. But in the diagram, $F_N = 45\ N$ is in the x - direction. Wait, maybe the labels are: $F_N$ (normal force, upward y - direction), $F_f$ (friction, left x - direction, $F_f=-30\ N$), $F_t$ (applied force, right x - direction), and $F_g$ (weight, downward y - direction).
     - For equilibrium in y - direction: $F_N=F_g$ (magnitude, opposite direction). If $F_N$ (normal force) has a magnitude of 45 N (but in the diagram, $F_N$ is in x - direction? Wait, maybe the diagram has a typo, and $F_N$ is in y - direction. Wait, the options: Let's look at the y - direction force ($F_g$) and x - direction force ($F_t$).
     - For the x - direction: The object is moving to the left, friction is to the right (opposing motion) or left? Wait, if the object is moving to the left, friction is to the right. But $F_f=-30\ N$ (left - ward), so the applied force $F_t$ (x - direction) should balance with $F_f$ and $F_N$? Wait, no, the options are about $F_t$ (x: $F_t$) and $F_g$ (y: $F_g$).
     - Let's consider the y - direction: The weight $F_g$ (downward) and the normal force (upward) should be equal in magnitude for equilibrium. If the normal force (in y - direction) has a magnitude equal to the weight. But in the diagram, the force with magnitude 45 N is in the x - direction. Wait, maybe the $F_N = 45\ N$ is the normal force in y - direction (upward), so the weight $F_g$ (downward) should be equal in magnitude, so $F_g=- 45\ N$ (if downward is negative y). But in the options, the y - direction force ( $F_g$): Let's check the options.
     - Option analysis:
       - Let's check the x - direction first. The friction force $F_f=-30\ N$ (left - ward). For equilibrium, the sum of x - direction forces: $F_N+F_f+F_t = 0$. If $F_N = 45\ N$ (right - ward, positive x), $F_f=-30\ N$ (left - ward, negative x), then $45-30 + F_t=0$, so $F_t=-15\ N$? No, that's not in the options. Wait, maybe the $F_N$ is a y - direction force. Let's assume that the upward is positive y, downward is negative y. The normal force $F_N = 45\ N$ (upward, positive y), so the weight $F_g$ (downward, negative y) should be $F_g=-45\ N$ (so magnitude 45 N, direction downward).
       - For the x - direction: The friction force $F_f=-30\ N$ (left - ward, negative x). The applied force $F_t$ (x - direction) should balance the friction force? Wait, no, if the object is in equilibrium, the sum of x - direction forces is zero. If there is only friction and the applied force in x - direction: $F_t+F_f = 0$, so $F_t=-F_f=30\ N$ (if $F_f=-30\ N$, then $F_t = 30\ N$ in the positive x - direction? But the object is moving to the left, so maybe the applied force is to the left? Wait, this is confusing. Wait, looking at the options:
         - Option 2: X: 30 N, Y: - 45 N. Let's see:
           - Y - direction: The weight $F_g$ is downward, so if upward is positive, $F_g=-45\ N$ (magnitude 45 N, direction downward).
           - X - direction: The friction force $F_f=-30\ N$ (left - ward). For equilibrium, the applied force $F_t = 30\ N$ (right - ward) to balance the friction? But the object is moving to the left, so maybe the friction is to the right? Wait, no, friction opposes motion. If the object is moving to the left, friction is to the right. So $F_f = 30\ N$ (right - ward), but in the diagram $F_f=-30\ N$ (left - ward). Maybe the coordinate system has left as positive x. Then, the object is moving to the left (positive x), friction is to the right (negative x), so $F_f=-30\ N$, and the applied force $F_t = 30\ N$ (positive x) to balance? No, net force would be $30-30 = 0$. And in y - direction, normal force (upward, negative y) and weight (downward, positive y)? No, this is getting too convoluted. Wait, the correct option should have $F_t$ (x - direction) with magnitude 30 N (to balance friction) and $F_g$ (y - direction) with magnitude 45 N (to balance normal force, assuming normal force is 45 N in y - direction). Wait, the option with X: 30 N and Y: - 45 N (if Y is upward, then weight is downward, so $F_g=-45\ N$) and X: 30 N (applied force to balance friction of 30 N) makes sense.
# Answer:
X: 30 N, Y: - 45 N (the option with X: 30 N and Y: - 45 N, which is the second option from the top: "X: 30 N, Y: - 45 N")

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