How Do You Calculate Tension in a Hammock's Ropes?

[stressedout09]

Homework Statement

A 51.0kg person takes a nap in a backyard hammock. Both ropes supporting the hammock are at an angle of 19.0 above the horizontal. Find the tension in the ropes.

Homework Equations

ty=tsin/theta
wy=-mg
possibly?

The Attempt at a Solution

tsin/theta+tsin/theta-mg=0
t=(51)(9.8)/2(sin19.0)
t=25...i know its wrong...someone help PLEASE, i don't understand anything my professor teaches so i can never apply anything. -_-

 

[method_man]

The Attempt at a Solution

tsin/theta+tsin/theta-mg=0
t=(51)(9.8)/2(sin19.0)
t=25...i know its wrong...

Perhaps you should use another bracket for the expression

t=(51)(9.8)/2(sin19.0)

t=(51)(9.8)/(2(sin19.0))

 

[tomcue]

As a scientist, it is important to approach problems with a clear and logical thought process. In this case, the problem requires finding the tension in the ropes supporting a hammock with a person of mass 51.0kg. To begin, we can draw a free body diagram to visualize the forces acting on the hammock and person.

From the diagram, we can see that the forces acting on the person are their weight (mg) and the tension in the ropes (ty). The vertical component of the tension in the ropes is ty*sin(theta), where theta is the angle of the ropes above the horizontal. We can also see that the person is at rest, so the sum of the forces in the vertical direction must equal zero.

Using this information, we can set up the following equation:

ty*sin(19.0) - mg = 0

Solving for ty, we get:

ty = mg/sin(19.0)

Plugging in the given values, we get:

ty = (51.0kg)(9.8m/s^2)/sin(19.0)

ty = 243.8N

Therefore, the tension in each rope is approximately 243.8N. It is important to note that this is the tension in each rope, as the weight of the person is evenly distributed between the two ropes.

In summary, the key to solving this problem is to identify the forces acting on the object and use the principles of equilibrium to set up an equation and solve for the unknown variable.