Two lifeguards pull on ropes attached to a
raft. If they pull in the same direction, the raft
experiences a net external force of 332 N to
the right. If they pull in opposite directions,
the raft experiences a net external force of
122 N to the left.
Draw a free body diagramfor each situation
and find the magnitude of the larger of the two
individual forces.
Answer in units of N
Resolve forces into x,y components.
Let the x-components of forces be f1,f2
then
f1+f2=332
f1-f2=122
Solving
f1=(332+122)/2=227
f2=(332-122)/2=105
Assuming θ=0 (in direction of raft)
so the forces are 227 and 105N
Sure, let's analyze the two situations and find the magnitude of the larger of the two individual forces.
1. When the lifeguards pull in the same direction (resulting in a net external force of 332 N to the right):
Free body diagram:
↑
------------------------------
| Raft |
------------------------------
↓
In this case, let's assume that the force exerted by one lifeguard is represented by F1, and the force exerted by the other lifeguard is represented by F2.
According to the problem, we have:
F1 + F2 = 332 N (right) ------- (Equation 1)
2. When the lifeguards pull in opposite directions (resulting in a net external force of 122 N to the left):
Free body diagram:
↑ ↑
------------------------------
| Raft |
------------------------------
↓ ↓
Using the same assumptions, we have:
F1 - F2 = 122 N (left) ------- (Equation 2)
To find the magnitude of the larger individual force, let's add these two equations together:
(Equation 1) + (Equation 2):
(F1 + F2) + (F1 - F2) = 332 N (right) + 122 N (left)
F1 - F2 cancels out because their signs are opposite, so we're left with:
2F1 = 454 N (right)
Dividing both sides by 2, we find:
F1 = 227 N (right)
Therefore, the magnitude of the larger of the two individual forces is 227 N.
In order to find the magnitude of the larger of the two individual forces, we need to analyze the forces involved in each situation separately.
When the lifeguards pull in the same direction, the net external force on the raft is 332 N to the right. In this scenario, we can draw the free body diagram for the raft as follows:
← F1 ← F2
-------------------------[RAFT]-----------------------
In this situation, we have two forces acting on the raft, both to the right. Let's call the magnitude of the larger force F1 and the magnitude of the smaller force F2. Since the net external force is to the right, F1 must be larger than F2.
Now, when the lifeguards pull in opposite directions, the net external force on the raft is 122 N to the left. Here is the free body diagram for this scenario:
F1 → ← F2
-------------------------[RAFT]-----------------------
In this situation, we still have two forces acting on the raft, but they are in opposite directions. The force F1 is to the right, and the force F2 is to the left. Since the net external force is directed to the left, we can conclude that F2 must be larger than F1.
Therefore, to find the magnitude of the larger of the two individual forces, we need to determine the value of F2. From the given information, we know that F2 = 122 N.
To solve this problem, we need to draw free-body diagrams for each situation and analyze the forces acting on the raft.
For the first situation where both lifeguards pull in the same direction and the net external force is 332 N to the right:
Free body diagram:
```
↑
│
F1 │ ------> 332 N
│
Raft │
```
From the diagram, we can see that there is only one force acting on the raft, which is the force exerted by one of the lifeguards (let's call it F1). This force has a magnitude of 332 N.
For the second situation where the lifeguards pull in opposite directions and the net external force is 122 N to the left:
Free body diagram:
```
↑
│
F1 │ ------> ?
│ │
Raft │ │
│ │ F2
│ │
↓
```
From the diagram, we can see that there are two forces acting on the raft - the force exerted by one lifeguard (F1) in one direction and the force exerted by the other lifeguard (F2) in the opposite direction. We are given that the net external force is 122 N to the left, so these two forces should cancel each other out.
The magnitude of the larger of the two individual forces can be found by realizing that the difference between the magnitudes of the two forces must be equal to the magnitude of the net external force. In this case, the magnitude of the larger force (F1 or F2) will be the sum of the net external force and the magnitude of the smaller force.
Let's assume F1 is the smaller force. Therefore, the magnitude of the larger force (F2) can be calculated as follows:
F2 - F1 = 122 N
Since the net external force to the left is 122 N, we can set this equal to the magnitude of the larger force:
F2 = F1 + 122 N
Now we have two equations:
F1 + F2 = 332 N (from the first situation)
F2 = F1 + 122 N (from the second situation)
By substituting the second equation into the first, we can solve for the magnitude of the larger force (F2):
F1 + (F1 + 122 N) = 332 N
Simplifying the equation:
2F1 = 210 N
Dividing both sides by 2:
F1 = 105 N
Since F2 = F1 + 122 N:
F2 = 105 N + 122 N
F2 = 227 N
Therefore, the magnitude of the larger of the two individual forces is 227 N.