Calculating Tension with a Centered Object

[ThePiGeek314]

Homework Statement

Two similar questions here. I have the same question regarding solutions to both.

Calculate the tension in a horizontal strand of spider web if a spider of mass 8.00 × 10⁻⁵ kg sits motionless in the middle of it. The strand sags at an angle of 12° below the horizontal.

A monkey (m = 4 kg) is in a harness connected to a rope that goes up over a pulley on the ceiling. If the monkey pulls on the other end of the rope, it will go up. If it is climbing at a constant velocity, what is the tension in the rope?

Homework Equations

F = ma
W = ga
T = ∑Fn of all objects hanging from the rope

The Attempt at a Solution

In the first problem, the gravitational force on the spider is 7.8 * 10^-4 N. Using right-triangle trig with the 12 degree given angle, the tension in the left half of the strand has to be 3.8 * 10^-3 N. But that's only one half of the strand.

Tension in left half = tension in right half...because the spider is sitting right in the center

So the total tension should be 3.8 * 2 * 10^-3 = 7.8 * 10^-3 N.

Answer key says 1.89 * 10^-3. 1.89 is a quarter of 7.8. Why is this? I know the spider is sitting in the middle of the web - I guess that means you might divide the 7.8 answer by 2, but not by 4.In the second problem, the gravitational force on the monkey is 39.2 N. The monkey is in equilibrium - moving with an acceleration of zero - so the tension on the rope must be 39.2 N.

But the correct answer is half that: 19.6 N. How can this be? Is it because the rope is part of a one-wheel pulley system, and that wheel divides the tension between the two rope segments? If it were a two-wheel system, would the tension then be 19.6 / 2 = 9.8 N?I think the answers to these questions are related. I can calculate the gravitational force, but I can't calculate tension with the force applied to the center of a rope, or in a pulley system. Can someone please explain this to me?

 

[BvU]

1) Spider is hanging from two strands, each of them pulling up. So one strand only has to exert 1/2 mg
2) same here: one section of the rope only has to lift half of Simmy weight.

If you have two scales and put one foot on each of them, you don't expect each scale to show double your body weight. Nor once. You expect the sum of the forces to be mg, nicely distributed over the two scales. Draw that FBD to see there is one mg down and therefore also just one mg up.

 

[ThePiGeek314]

Is this why each of the ropes in a single pulley only has tension equal to HALF the weight in the entire pulley system?

 

[CWatters]

Yes.

The approach is to recognise that the spider isn't accelerating so Newton says the net force acting on him is zero (in any direction, eg vertically or horizontally). Then draw the Free Body Diagram for the spider.

If you take upwards as +ve then summing in the vertical direction you have..

+T*Sin(12) +T*Sin(12) - mg = 0
or
2T*Sin(12) - mg = 0

where
T is the tension in each strand of web
m is the mass of spider
rearrange to give
T = mg/(2*Sin(12)

For completeness.. summing horizontally you would get
T*Cos(12) - T*Cos(12) = 0

 

[BvU]

Hello $\Pi G$,

Is this why each of the ropes in a single pulley only has tension equal to HALF the weight in the entire pulley system?

Yes.

By now your helpers think it necessary you learn how to draw a free body diagram for an object (e.g. a pulley):
Simply draw all the forces that act on one object

Example: massless pulley with massless rope, from each end of which hangs a mass of 10 kg :

Situation sketch on the left, FBD on the right. Only way the pulley can stay in equilibrium (i.e. not accelerate in any direction) is if F_susp = 20 g Newton. That is the upwards force the pulley exerts on the suspension.

Likewise you can draw FBDs for the left mass, right mass, blue suspension bar, ceiling, rope, etc.

For your spider the situation is different, but not too much. Can you draw a free body diagram and post it ?
Idem monkey.
And superhero.

Learning to do this kind of stuff properly is a very good investment.

 

[ThePiGeek314]

I'm pretty confident with drawing free-body diagrams. I just posted one from the Internet because it would have been more hassle to draw my own, take a picture, upload it, and post it.

 

[BvU]

OK, good. Carry on !