Educational worksheet on balanced and unbalanced forces featuring various scenarios with force vectors.

Let's go through each diagram one by one, analyze the forces, and answer the questions:

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Diagram 1: Bird on a Stand

Forces:
- Downward arrow: 500 N → Weight (Gravity)
- Upward arrow: 500 N → Normal Force (Support from stand)

Analysis:
- The forces are equal in magnitude (500 N) but opposite in direction.
- Net force = 0 N → Balanced forces

Answers:
a. Balanced
b. Remain stationary

> ✔ Explanation: Since the upward normal force equals the downward gravitational force, the bird doesn't accelerate — it stays at rest.

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Diagram 2: Woman on a Seesaw

Forces:
- Left side: 70 N down → Weight of object/person on left
- Right side: 60 N down → Weight of object/person on right
- Middle: 120 N up → Normal force from pivot (support)

Analysis:
- Total downward force = 70 N + 60 N = 130 N
- Upward force = 120 N
- Net force = 130 N down – 120 N up = 10 N downward → Unbalanced
- However, since this is a seesaw, we must consider torque, not just net force. But the question asks about overall forces, so let’s focus on vertical forces only.

But wait: The 120 N upward force is not enough to balance 130 N downward, so there is a net downward force of 10 N, meaning the system is unbalanced vertically.

However, in real life, a seesaw would rotate due to unequal torques, but here the forces are labeled vertically.

Still, if the pivot can’t support more than 120 N, the seesaw will move downward where the total force is greater.

But note: the pivot is likely supporting the whole structure. Let's assume the 120 N upward force is the normal reaction from the ground.

So:
- Total downward = 70 + 60 = 130 N
- Upward = 120 N
- Net force = 10 N downward → Unbalanced

But that implies the entire seesaw is accelerating downward? That seems odd because the pivot should be fixed.

Actually, the pivot provides a reaction force. If the total weight is 130 N, then the pivot must provide 130 N upward to keep it balanced. Here, it shows only 120 N, which means the system is unbalanced, and the seesaw might sink or break.

But assuming the diagram is correct as drawn:

Answer:
a. Unbalanced
b. Will move downward (due to excess downward force)

> ⚠️ But in reality, a seesaw pivots, so motion depends on torque. However, based on vertical forces only, since 130 N > 120 N, net downward force exists.

But let’s reconsider: maybe the 120 N is the total upward support, and 130 N downward exceeds it → unbalanced, so the seesaw would accelerate downward unless supported.

But typically, the pivot supports the full weight. So perhaps this is an error?

Wait — actually, the 120 N upward is likely the force from the pivot, and the two downward forces are weights.

So:
- Total downward = 70 + 60 = 130 N
- Upward = 120 N
→ Net downward force = 10 N → Unbalanced

So:
a. Unbalanced
b. Will move downward

> ✔ But in practice, the pivot would exert more force to balance. This may indicate an unstable situation.

We’ll go with what’s shown.

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Diagram 3: Parachutist

Forces:
- Downward: 500 N → Weight (Gravity)
- Upward: 300 N → Air Resistance (Drag)

Analysis:
- Downward force > Upward force → Net downward force
- Unbalanced forces → Acceleration downward

Answers:
a. Unbalanced
b. Will move downward

> ✔ Explanation: Gravity pulls the person down harder than air resistance pushes up, so they accelerate downward.

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Diagram 4: Cannon

Forces:
- Right: 1600 N → Thrust (from explosion)
- Left: 15 N → Air resistance / friction?
- Down: 45 N → Weight (Gravity)

Wait — the 15 N is labeled left, and 1600 N right, so horizontal forces:
- Right: 1600 N
- Left: 15 N
→ Net horizontal force = 1600 – 15 = 1585 N to the right

Vertical:
- Down: 45 N (weight)
- No upward force shown → probably implied that it’s on ground, so normal force balances weight

But only downward force shown → missing normal force from ground?

But the question says "label the arrows" — so we label only those shown.

But for analysis, we need to know all forces.

But only three forces shown: thrust (1600 N →), air resistance (15 N ←), gravity (45 N ↓)

No upward force → so vertical forces unbalanced? But cannon is on wheels, so normal force exists, but not shown.

But the horizontal forces are clearly unbalanced.

So:
- Horizontal net force = 1600 – 15 = 1585 N to the right
- Vertical: weight 45 N down, no upward arrow → but we assume ground supports it (normal force not shown)

But since the question asks about overall forces, and only these are shown, we must consider only what’s visible.

But the motion depends on net force.

Even if normal force is present, the horizontal force is large and unopposed except for 15 N.

So:
a. Unbalanced
b. Will move to the right

> ✔ Because thrust (1600 N) >> air resistance (15 N), so net force right → accelerates right.

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Diagram 5: Two Cars Colliding

Forces:
- Left car: 1100 N → (pushing right)
- Right car: 1400 N ← (pushing left)

But these are forces acting on the cars — likely contact forces between them.

So:
- Left car experiences 1100 N to the right (from the right car)
- Right car experiences 1400 N to the left (from the left car)

But Newton’s Third Law: action-reaction pairs must be equal and opposite.

But here, the forces are not equal — 1100 vs 1400.

This suggests the diagram shows the forces acting on each car, not the pair.

So:
- Net force on left car: 1100 N → (to the right)
- Net force on right car: 1400 N ← (to the left)

But they are pushing on each other — so the forces should be equal and opposite.

Unless external forces are involved.

But the labels suggest:
- Left car has a force of 1100 N to the right
- Right car has a force of 1400 N to the left

So they are not equal, which violates Newton’s Third Law.

But perhaps the numbers are the net forces on each car, and the interaction is part of it.

But regardless, each car has a net force.

But the question is: are the overall forces acting on the system balanced?

But we’re analyzing each object separately.

Looking at the diagram: two cars connected or touching.

But the forces shown are:
- On left car: 1100 N →
- On right car: 1400 N ←

So the forces are unbalanced — net force on the system is 1400 – 1100 = 300 N to the left

So:
a. Unbalanced
b. Will move to the left

> ✔ Because the total force on the system is 300 N to the left (since 1400 N left > 1100 N right), so the system accelerates left.

But again, if they are touching, the internal forces cancel, and net force is from external sources.

But here, both forces are shown as acting on the cars — possibly external forces.

So yes, net force = 1400 N left – 1100 N right = 300 N left

So:
a. Unbalanced
b. Move to the left

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Diagram 6: Snowboarder

Forces:
- Down: 115 N → Weight (Gravity)
- Up: 115 N → Normal force (from snow)
- Left: 135 N → Friction / Air resistance?
- Right: 25 N → Push from snow? Or component of motion?

Wait — likely:
- 115 N down → Weight
- 115 N up → Normal force
- 135 N left → Friction (resisting motion)
- 25 N right → Possibly push from snow, or something else?

But if the snowboarder is moving, and there’s a 25 N to the right, but 135 N to the left, then net horizontal force = 25 – 135 = –110 N → left

So net force to the left → decelerating or moving left?

But the labels say:
- 115 N ↓ → Weight
- 115 N ↑ → Normal force → balanced vertically
- 135 N ← → Friction
- 25 N → → Push? (maybe from snow or slope)

But if 25 N to the right and 135 N to the left, net horizontal force = 110 N to the left

So:
a. Unbalanced
b. Will move to the left

But if she was going right, she slows down; if going left, speeds up.

But the question says “which direction the object will move” — but if already moving, it could be changing speed.

But assuming it starts from rest, it will move in the direction of net force.

Net horizontal force is 110 N to the left, so:
- a. Unbalanced
- b. Will move to the left

But wait — is the 25 N a forward push? Maybe it's the component of gravity down the slope?

But the diagram shows:
- 115 N down (gravity)
- 115 N up (normal)
- 135 N left (friction)
- 25 N right

So unless there’s a slope, gravity is straight down.

But if the snowboarder is on a slope, the normal force might not be exactly up.

But here, normal force is shown directly up, same as gravity → implies flat surface.

So if on flat ground, and gravity and normal force balance, then only horizontal forces matter.

Net horizontal force = 25 N right – 135 N left = 110 N left

So:
a. Unbalanced
b. Will move to the left

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Final Answers:

#### Diagram 1: Bird on Stand
- Arrows:
- ↓ 500 N → Weight (Gravity)
- ↑ 500 N → Normal Force
- a. Balanced
- b. Remain stationary

#### Diagram 2: Seesaw
- Arrows:
- ↓ 70 N → Weight (Left side)
- ↓ 60 N → Weight (Right side)
- ↑ 120 N → Normal Force (Pivot support)
- a. Unbalanced
- b. Will move downward (due to excess downward force)

#### Diagram 3: Parachutist
- Arrows:
- ↓ 500 N → Weight (Gravity)
- ↑ 300 N → Air Resistance (Drag)
- a. Unbalanced
- b. Will move downward

#### Diagram 4: Cannon
- Arrows:
- → 1600 N → Thrust (Explosion)
- ← 15 N → Air Resistance
- ↓ 45 N → Weight (Gravity)
- a. Unbalanced
- b. Will move to the right

#### Diagram 5: Two Cars
- Arrows:
- → 1100 N → Force on left car (from right car)
- ← 1400 N → Force on right car (from left car)
- a. Unbalanced
- b. Will move to the left (net force left)

#### Diagram 6: Snowboarder
- Arrows:
- ↓ 115 N → Weight (Gravity)
- ↑ 115 N → Normal Force
- ← 135 N → Friction
- → 25 N → Push (or applied force)
- a. Unbalanced
- b. Will move to the left

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Summary Table:

| Diagram | a. Balanced/Unbalanced | b. Direction/Motion | Labeled Forces |
|--------|------------------------|----------------------|----------------|
| 1 | Balanced | Remain stationary | Weight, Normal Force |
| 2 | Unbalanced | Move downward | Weight (left), Weight (right), Normal Force |
| 3 | Unbalanced | Move downward | Weight, Air Resistance |
| 4 | Unbalanced | Move to the right | Thrust, Air Resistance, Weight |
| 5 | Unbalanced | Move to the left | Force on left car, Force on right car |
| 6 | Unbalanced | Move to the left | Weight, Normal Force, Friction, Applied Force |

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✔ Final Answer Key:

Diagram 1:
a. Balanced
b. Remain stationary

Diagram 2:
a. Unbalanced
b. Will move downward

Diagram 3:
a. Unbalanced
b. Will move downward

Diagram 4:
a. Unbalanced
b. Will move to the right

Diagram 5:
a. Unbalanced
b. Will move to the left

Diagram 6:
a. Unbalanced
b. Will move to the left

> All forces labeled as per context (Weight, Normal Force, Air Resistance, Thrust, Friction, etc.) as shown in diagrams.