A steel wire 40 m long is suspended between two fixed points 20 m apart.

A force of 375 N pulls the wire down at a point 15 m from one end of the
wire. State the tension in each part of the wire.

Texbook answer: 375N, 0N

Well, let me give you a different perspective on this.

So, we have a steel wire that's 40 meters long and suspended between two fixed points 20 meters apart. Now, imagine this wire as a high wire act in a circus, with the force of 375 N pulling it down at a point 15 m from one end.

Now, let's consider the tension in each part of the wire. We can clearly see that the force of 375 N is only acting on one side of the wire. That poor wire is feeling all the tension. It's like when you're carrying all the grocery bags by yourself while your friends are just standing there watching you struggle. So, in this case, the tension in that part of the wire is 375 N.

But what about the other side of the wire? Well, since nobody is pulling it down, it's just hanging out there, relaxing. It's like the person who got the day off while everyone else is working. So, in this case, the tension in that part of the wire is 0 N. Zero tension, just chillin'.

So, to sum it up, we have 375 N of tension in one part of the wire and 0 N of tension in the other part. It's an unequal distribution of tension, but hey, life is never fair, even for steel wires.

Draw the position diagram, showing the actual suspension

I have a triangle PAQ, so that PQ = 20, PA = 15 and AQ = 25
notice that 15:20:25 = 3:4:5
so the triangle actually is right-angled, with angle P = 90°
Thus PA is vertical, as long as PQ is horizontal, and all the tension
will be in string PA
So the tension in PA is 375N and the tension in string AQ will be 0 N.

Not sure if the author of the question had that in mind, since it makes the
question rather trivial.

To find the tension in each part of the wire, we can first analyze the forces acting on the wire.

At the point where the force is applied, 15 m from one end of the wire, there are two forces acting on it:

1. Tension force towards the left end of the wire.
2. Tension force towards the right end of the wire.

Since the wire is in equilibrium, these two tension forces must be equal in magnitude. Let's call the tension force towards the left end of the wire T1 and the tension force towards the right end T2.

Considering the forces acting on the wire at the 15 m point:

1. Tension force T1 towards the left end of the wire.
2. Tension force T2 towards the right end of the wire.
3. The force of 375 N pulling the wire down.

The tension force T1 and T2 will act horizontally along the wire, while the 375 N force will act vertically downward. These three forces must balance each other out for the wire to be in equilibrium.

Since the wire is symmetric, we can conclude that T1 = T2, meaning the tension in each part of the wire is the same.

Therefore, based on the given information and the principles of equilibrium, the tension in each part of the wire is:
Tension towards the left end (T1) = Tension towards the right end (T2) = 375 N.

To find the tension in each part of the wire, we can consider the forces acting on the wire. First, let's divide the wire into two sections: the left part and the right part.

In the left part of the wire, we have a force of 375 N pulling it down. Since the wire is in equilibrium (not moving), the net force acting on it must be zero. So, at the point 15 m from one end, there must be an upward force (tension) that balances the 375 N downward force. Let's call this tension T1.

Now, let's consider the right part of the wire. Since there are no forces acting on this part, the tension in this part of the wire must be zero, as stated in the textbook answer. Let's call this tension T2.

Therefore, the tensions in each part of the wire are T1 = 375 N (upward at the point 15 m from one end) and T2 = 0 N (no tension in the right part).

By considering the forces and equilibrium conditions, we can determine the tensions in different parts of the wire.