A 240-N weight is hung from two ropes as shown. The tension in the horizontal rope has magnitude:
A) 0
B) 656 N
C) 480 N
D) 416 N
E) 176 N

Question

A 240-N weight is hung from two ropes as shown. The tension in the horizontal rope has magnitude: 
A) 0 
B) 656 N 
C) 480 N 
D) 416 N 
E) 176 N

Quizplus · Accepted Answer

We can start by analyzing the forces acting on the weight. There are two ropes, one vertical and one horizontal, and the weight itself. The weight is acted upon by its own weight force of 240 N, acting downwards.

The vertical rope is supporting the weight completely, so it must be exerting a force equal to the weight force upwards. This force is equal to 240 N.

The horizontal rope is preventing the weight from moving horizontally, so it must be exerting a force in the opposite direction. We don't know the magnitude of this force yet, so let's call it T.

We can now use trigonometry to relate the horizontal and vertical forces to the tension force T. Let's call the angle between the horizontal rope and the vertical rope theta. From the diagram, we can see that sin(theta) = 3/5 and cos(theta) = 4/5.

The horizontal force can be found using the following equation:
horizontal force = T*cos(theta)

The vertical force can be found using the following equation:
vertical force = T*sin(theta) + weight force

We know that the vertical force must equal the weight force, so we can set these two equations equal to each other and solve for T:

T*sin(theta) + weight force = weight force
T*sin(theta) = 0
T = 0 (since sin(theta) cannot be zero)

This means that there is no tension in the horizontal rope, which contradicts our assumption that the weight is being prevented from moving horizontally. So, we made a mistake somewhere.

Let's try again, but this time let's assume that there is tension in the horizontal rope. We can repeat the same calculations as before, but this time with a non-zero value for T:

horizontal force = T*cos(theta) = T*(4/5)
vertical force = T*sin(theta) + weight force = T*(3/5) + 240

Again, we know that the vertical force must equal the weight force, so we can set these two equations equal to each other and solve for T:

T*(3/5) + 240 = 240
T*(3/5) = 0
T = 0 (since sin(theta) cannot be zero)

This is the same result as before, which means that our assumption that there is tension in the horizontal rope was incorrect. So, we must have made another mistake.

The mistake is that we assumed that the weight is in equilibrium when it is not. The weight is being pulled downwards by gravity, so it is accelerating downwards. This means that the net force acting on the weight is not zero, so we cannot use the equations for equilibrium that we were using before.

We can still use trigonometry to find the tension force in the horizontal rope, but we need to take into account the acceleration of the weight. Let's call the acceleration of the weight a.

The horizontal force can still be found using the equation:
horizontal force = T*cos(theta)

The vertical force now needs to include the weight force and the force of gravity creating the acceleration of the weight. The force of gravity is given by:
force of gravity = mass * acceleration due to gravity (g)

The mass of the weight can be found using the equation:
mass = weight force / acceleration due to gravity = 240 / 9.8

The force of gravity can then be found using the equation:
force of gravity = mass * acceleration due to gravity = (240 / 9.8) * 9.8 = 240

The vertical force now becomes:
vertical force = T*sin(theta) + weight force + force of gravity

The net force acting on the weight is equal to its mass times its acceleration, or:
net force = mass * acceleration = 240 * a

The net force can also be calculated as the difference between the vertical force and the weight force:
net force = vertical force - weight force = T*sin(theta) + 240 - 240 = T*sin(theta)

Setting these two equations equal to each other, we can solve for T:

240 * a = T*sin(theta)
T = (240 * a) / sin(theta)

We don't know the value of the acceleration, so we need to find it. The weight is accelerating downwards, so its acceleration is equal to the acceleration due to gravity, which is approximately 9.8 m/s².

Substituting this value into the equation for T, we get:
T = (240 * 9.8) / sin(theta) = 2352 / 3

Rounding to the nearest integer, we get T = 784 N. However, this is not one of the answer choices.

We made one final mistake, which is that we forgot to take into account the fact that the weight is being supported by two ropes, not one. The tension in the vertical rope is equal to the weight force, or 240 N. This means that the tension in the horizontal rope must be less than 240 N, since the weight is not being supported completely by the horizontal rope.

Looking at the answer choices, the only one that is less than 240 N is D) 416 N. Therefore, the correct answer is:

Answer: D
 The tension in the horizontal rope has a magnitude of 416 N.

A 240-N Weight Is Hung from Two Ropes as Shown