A rope 26 ft long is attached to two points A and B, 20 ft apart at the same level. A load of 250 lb is carried at the middle of the rope. What forced is exerted on A?
The rope hangs in a "V" shape. The two sides of the rope make angles of
cos^-1 (10/13) = 39.7 degrees with the horizontal. Each side of the rope suppoerts half the weight. If T is he tension force in the rope,
2 T sin39.7 = (weight) = 250 lb
T = 125/sin39.7 = 195.6 lb
To find the force exerted on point A, we need to consider the equilibrium of forces acting on the system.
Given:
- Total length of the rope (L) = 26 ft
- Distance between points A and B (d) = 20 ft
- Load at the middle of the rope (W) = 250 lb
In this scenario, we can assume that the rope is massless, and all forces are acting vertically.
We'll break down the problem into two parts:
1. Forces due to the weight of the load:
Since the load is at the middle of the rope, the weight (W) will be equally distributed on both sides. Therefore, the load on each side is 250 lb / 2 = 125 lb.
2. Forces due to the tension in the rope:
The tension in the rope will be different on each side since the load is not at the center of the rope.
To calculate the tension on the left side (TA) and the right side (TB) of the rope, we can use the concept of equilibrium. At the points A and B, the vertical forces must balance each other.
Let's consider the left side of the rope (TA):
- The downward force due to the weight of the load (125 lb)
- The upward force due to the tension in the rope (TA)
- There is no other vertical force acting on this part of the rope
Since the system is in equilibrium, the sum of all vertical forces must be zero:
125 lb - TA = 0
Now let's consider the right side of the rope (TB):
- The downward force due to the weight of the load (125 lb)
- The upward force due to the tension in the rope (TB)
- The downward force due to the weight of the rope itself
Since the total length of the rope (L) is 26 ft and the distance between A and B (d) is 20 ft, the length of the rope on each side is (L - d)/2 = (26 ft - 20 ft)/2 = 6 ft. Therefore, the weight of the rope itself is 6 ft / L * W = 6/26 * 250 lb = 57.69 lb (approximately).
Applying the concept of equilibrium, the sum of all vertical forces must be zero:
125 lb + 57.69 lb - TB = 0
Now we have two equations:
125 lb - TA = 0
125 lb + 57.69 lb - TB = 0
Solving these equations simultaneously will give us the tension forces:
TA = 125 lb
TB = 182.69 lb
Finally, to find the force exerted on point A, we need to consider that A is being pulled downward by the tension force:
Force on A = TA - Weight of the rope = 125 lb - 57.69 lb = 67.31 lb
Therefore, the force exerted on point A is approximately 67.31 lb.
To find the force exerted on point A, we can consider the forces acting on the load and then apply Newton's third law of motion.
1. Since the load is at the middle of the rope, we can assume it is evenly distributed. Therefore, the weight of the load is shared equally on both sides of the middle point.
Weight of load on each side = (250 lb) / 2 = 125 lb.
2. Now let's consider the forces acting on the left side of the rope.
At point A, there are two forces:
- The upward force exerted by the load on the left side, which is 125 lb.
- The tension force in the rope, pulling to the right.
3. By applying Newton's third law of motion, we know that for every action, there is an equal and opposite reaction. Therefore, the tension force pulling to the right at point A is also exerted on point B.
4. Since the rope is attached at point B and the distance between A and B is 20 ft, the tension force exerted on B can be calculated using the concept of torque.
Torque = force × perpendicular distance
Torque exerted on B = Tension force (at B) × Distance (from A to B)
Torque exerted on B = Tension force (at B) × 20 ft
5. Since the rope is in equilibrium (not rotating or accelerating), the torque at point A must be zero.
Torque exerted on A = 0
6. The torque exerted on B (as calculated in step 4) is equal to the torque exerted on A (as calculated in step 5).
Therefore, we can write the equation:
Tension force (at B) × 20 ft = 0
7. Since the product of the tension force at point B and the distance from A to B is zero, either the tension force is zero or the distance is zero.
However, the distance cannot be zero because point B is 20 ft away from point A.
Thus, the tension force at point B must be zero.
8. Using Newton's third law, we know that the tension at point A and the tension at point B are equal and opposite forces.
Therefore, the force exerted on point A is also zero.
In conclusion, there is no force exerted on point A.