A weight of 50.0 N is hung from the center of a stretched horizontal wire 10.0 meters long. It causes the wire to say 1.00m at its center. What is the force exerted by each half of the wire in supporting the weight?
1. due to symettry, assume each half is supporting 25N
So you are looking for tension.
SinTheta=1m/5m based on lengths
SinTheta=25N/tension based of forces
25/Tension=1/5
tension= 125N
check that.
Thank you so so much And I also posted another physics question down below if you don't mind helping me out?
To find the force exerted by each half of the wire in supporting the weight, you can use the concept of tension in a string or wire. Here's how you can approach this problem:
Step 1: Understand the Setup
Consider the wire to be divided into two halves symmetrically. The weight is hung from the center of the wire, so the tension will be the same on both halves.
Step 2: Identify the Relevant Forces
There are two main forces acting on the wire: the weight hanging from the center and the tension in the wire.
Step 3: Determine the Total Tension
The total tension in the wire is equal to the weight. In this case, the weight is 50.0 N.
Step 4: Split the Tension
Since the wire is symmetrically divided into two halves, the tension will be equally distributed between the halves when the wire is in equilibrium.
Step 5: Calculate the Force Exerted by Each Half
To find the force exerted by each half of the wire, divide the total tension by 2. Therefore, the force exerted by each half of the wire is 50.0 N / 2 = 25.0 N.
So, the force exerted by each half of the wire in supporting the weight is 25.0 N.