Calculating Tension of Threads for Suspended Spider

[faoltaem]

hey i have a question from my physics tute but I'm lost.

a spider is suspended on lines of web between two level branches. each line of web is 72cm long and the branches they are attached to are 1.1m apart
1) given that the spider weighs 5.2g, calculate the tension force in either of the 72cm threads.
2) a gust of wind blows the branches closer together. does the tension in the threads increase, decrease or remain the same?

i'm normally pretty good at mathmatic type questions but I'm just really lost,like i get that the weight needs to become
W = mg = 0.0052 x 9.81 = 0.051 N
but i think I'm lost because I'm used to working with angles and because the triangle formed doesn't have any right angles i can't work any out.
so even if you could just give me a formula i'd be really grateful, but any help you can give me would be great.

 

[Astronuc]

W = mg = 0.0052 x 9.81 = 0.051 N

So the braches are 1.1 m apart (and level), and the threads are each 0.72 m long.

So form a triangle 1.1 m horizontal base and two legs of 0.72 m. The spider is positioned below the midpoint of the 1.1 m separation.

 

[faoltaem]

my question came with a picture so i get how it's all set out I'm just having a problem with how to calculate the tension of the threads (.72m) leading down to the spider.
I'd be abe to work it out if i had a formula for it, but I've looked through my textbook, my lecture notes and had a quick look online but wasn't able to find a formula that didn't have either an angle or a velocity at which the spider would be traveling (except gravity of course).

 

[Astronuc]

This is a statics problem - "a spider is suspended". Gravity pulls vertically (straight) down.

Both threads bear the weight of the spider. Given the symmetry, they each bear have the weight.

The threads are oriented at angle with the horizontal (with the complementary angle with respect to the vertical), so there are horizontal and vertical components in the tension.

The vertical component of tension must relate to the weight of the spider.

The horizontal component of tension arises since one thread must pull the other thread.

If one has a right triangle with a horizontal leg of 0.55 m and a hypotenuse of 0.72 m, what is the length of the vertical leg? What is the angle of the hypotenuse with respect to the horizontal?

 

[faoltaem]

thanks that helped heaps, this is what I've got now, just wanted to check that I've done the right thing

a$$^{2}$$ + b$$^{2}$$ = c$$^{2}$$
b = $$\sqrt{c^{2} - a^{2}}$$
= $$\sqrt{0.72^{2} - 0.55^{2}}$$
= 0.46m

cos $$\theta$$ = $$\frac{A}{H}$$

cos $$\theta$$ = $$\frac{0.55}{0.72}$$

$$\theta$$ = cos$$^{-1}$$ $$\frac{0.55}{0.72}$$
= 40$$\circ$$

T$$_{y}$$ = T sin $$\theta$$
W$$_{y}$$ = -mg
$$\sum$$ F$$_{y}$$ = 0 = 2T$$_{y}$$ + W$$_{y}$$
= 2(T sin $$\theta$$) - mg
2(T sin $$\theta$$) = mg
T sin $$\theta$$ = $$\frac{mg}{2}$$
T = $$\frac{mg}{2 sin \theta}$$

= $$\frac{0.0052 x 9.81}{2 sin 40\circ}$$

= 0.0395N.

and not neglecting the second part the tension would decrease as the branches got closer together

 

[Astronuc]

The T^y component acts vertically, so the angle between it and the hypotenuse is the angle with respect to the vertical. Make sure you solve for the correct angle.

 

[faoltaem]

ok so that would make the angle 50$$\circ$$

so T = $$\frac{mg}{2sin\theta}$$

= $$\frac{0.0052 \times 9.81}{2 sin 50}$$

= 0.033N

= 3.3 x 10$$^{-2}$$N

 

[Astronuc]

a$$^{2}$$ + b$$^{2}$$ = c$$^{2}$$
b = $$\sqrt{c^{2} - a^{2}}$$
= $$\sqrt{0.72^{2} - 0.55^{2}}$$
= 0.46m

Correct.

But

sin $$\theta$$ = $$\frac{A}{H}$$, where A is the horizontal leg, if $$\theta$$ is angle with vertical

or cos $$\theta$$ = $$\frac{B}{H}$$, where B is the vertical leg

Thus,
sin $$\theta$$ = $$\frac{0.55}{0.72}$$

Then
$$\theta$$ = sin$$^{-1}$$ $$\frac{0.55}{0.72}$$
= 50$$\circ$$,

where $$\theta$$ is the angle with respect to vertical.

Then
T$$_{y}$$ = T cos $$\theta$$
W$$_{y}$$ = -mg
$$\sum$$ F$$_{y}$$ = 0 = 2T$$_{y}$$ + W$$_{y}$$
= 2(T cos $$\theta$$) - mg
2(T cos $$\theta$$) = mg
T cos $$\theta$$ = $$\frac{mg}{2}$$
T = $$\frac{mg}{2 cos \theta}$$

If one uses the angle with respect to horizontal, then one must switch sin and cos.


I just wanted to make sure you used the correct angle, which you did in post #5. Sorry for the confusion.