A rope, assumed massless, is stretched horizontally between two supports that are

3.44 m apart. When an object of weight 3160 N is hung at the center of the rope,

the rope is observed to sag by 35 cm. Calculate the tension in the rope.

# A rope, assumed massless, is stretched horizontally between two supports that are…

Complete Question Text:*AP Physics + 19 others*

*Aerospace Engineering*

The first step in solving any physics problem like this is to draw out the force diagram with correctly drawn vectors indicating direction and magnitude of each force. In this case, we have three forces acting on the object being hung at the center of the rope, as you can see in the diagram below.

A quick note here: The angle on either side of the rope is alpha because the system is symmetric, since it is given that the object is hung at the center of the rope. Also note that since the rope is assumed massless, tension is uniform throughout the rope and the rope exerts two forces of equal magnitude on both the left and right side of the object.

The next step is to choose the right coordinate system with which to analyze this object. The simplest to set up here would probably be a Cartesian coordinate system with i hat (the positive x direction) pointing to the right, and j hat (the positive y direction) pointing straight up.

The next step to do then is to break up each force into i and j hat components. It is quite easy to do this with the weight of the object, because gravity only points downwards in the negative j hat direction. However to find the horizontal and vertical components of the two tension forces on either side, we must first know the angle alpha.

Solving for alpha is done by taking the inverse tangent of the sag length (s = 35 cm), divided by half of the distance between the supports (1.77 m = 3.44/2).

The whole expression is arctan(0.35 m/ 1.44 m)

Plugging this into a calculator will easily give you alpha.

Now that alpha is accounted for we can split up the tension forces using trigonometry. The horizontal components are always T*cos(alpha) and the vertical components are T*sin(alpha).

The decomposition of these vectors allows us to establish two equations, one for each direction of motion. However because we only have unknown (T), then we really only need to use one equation, which is the equation in the j-hat (vertical) direction.

The equation to start out with is F = m*a.

The left hand side consists of the vertical components of the two tensions, both pointing in the positive j-hat direction, and the weight of the object pointing in the negative j-hat direction. The sum of these forces is:

T*sin(alpha) + T*sin(alpha) + (-3160 N)

Note: N in this case is just standing for units of Newtons, not another variable.

On the right hand side, the fact that the object is in static equilibrium tells us that the acceleration in the vertical direction is 0, and thus the entire right hand side of this equation goes to 0.

Simplifying the first expression leads to this: 2*T*sin(alpha) - 3160 N

Merging this with the right hand side gives us a simple algebraic expression for T that can be solved easily.

2*T*sin(alpha) - 3160 N = 0

Which solving for T gives you: T = (3160 N) / [2*sin(alpha)]

Plugging in for alpha gives you your final equation.

T = (3160 N) / [2*sin(arctan(0.35 m/ 1.44 m))] which is approximately 6690 N.

## The student's feedback

Draw a force diagram for the weight. The tension force is pulling on the weight in two directions, along the rope. Their horizontal components cancel out, and their vertical components sum together to counteract the weight. So you get 2*T_vertical = mg. (Both tension forces are the same because it's the same rope and tension is the same everywhere in a massless rope). So you can solve for T_vertical and then use that to figure out T, knowing the angle of the rope (you'll have to do some trig on the distances they give you). That looks something like T_vertical = T sin(theta) for some angle theta; work it out from your diagram.

Let me know if you need any more help!