Find the tension in the wire and the components of the force exerted

In summary: I stated in one of my other posts that the numbers were exactly what my actual problem contained. the 720 N of the bear is actually 740, which is why when I added all the numbers up I got 1020 instead of 1000.
  • #1
yb1013
56
0

Homework Statement



A hungry bear weighing 720 N walks out on a beam in an attempt to retrieve a basket of food hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 5.40 m long; the basket weighs 80.0 N.

a) When the bear is at x = 1.00 m, find the tension in the wire and the components of the force exerted by the wall on the left end of the beam.
N (tension)
N ( Fx)
N ( Fy)

The Attempt at a Solution



Okay well I sort of understood this question and created a balanced torque equation. After doing this I got that:

Tension = 345.66
Fx = 172.83

But I need help with finiding Fy Please.
I only have like one more chance to answer the question on my homework so I want to make sure I can get the right answer.
I know I have to write out some equation like maybe:

Rsin(theta) = 720 + 200 + 80 - 345.66sin(60)

Is this correct? And if its not, am I on the right track?
 
Last edited:
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  • #2
is the beam supported by a horizontal string on each end?
 
  • #3
Need more data on how the beam is supported.

Sounds like a cantilever beam with a tension wire at the other end, but is this tension wire vertical or anchored to the wall.
 
  • #5
yb1013 said:
A hungry bear weighing 720 N walks out on a beam in an attempt to retrieve a basket of food hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 5.40 m long; the basket weighs 80.0 N.

a) When the bear is at x = 1.00 m, find the tension in the wire and the components of the force exerted by the wall on the left end of the beam.
…
Tension = 345.66
Fx = 172.83

But when I go for Fy using 345sin(60), I keep getting 299 which my homework keeps telling me is wrong..

Hi yb1013! :smile:

I think you've calculated the y-component of the tension, instead of the y-component of the force at the wall. :redface:
 
  • #7
would this be closer to the y-component for the force?

Rsin(theta) = 720 + 200 + 80 - 345.66sin(60)
 
  • #8
what a dumb bear...

Anyways, I broke the componets into x and y directions. The y dir. is Tcos(theta) -mg =0 due to their being no motion.

also, their are 3 masses. bear, basket, and beam.
 
  • #9
xxChrisxx said:
Its yogi bear...!

Thats some nice pic-a-nic

He's smarter than the average bear! :biggrin:
yb1013 said:
would this be closer to the y-component for the force?

Rsin(theta) = 720 + 200 + 80 - 345.66sin(60)

Yup … if the tension is right, then that looks right too! :smile:
 
  • #10
Ashleyz said:
what a dumb bear...

Anyways, I broke the componets into x and y directions. The y dir. is Tcos(theta) -mg =0 due to their being no motion.

also, their are 3 masses. bear, basket, and beam.


So would it go like 345.66cos(60) - 1020 = 0 ??
 
  • #11
tiny-tim said:
He's smarter than the average bear! :biggrin:


Yup … if the tension is right, then that looks right too! :smile:

Ok well the numbers in that question aren't actually my exact numbers, but with my numbers I come out with:

Rsin(theta) = 720.65

would that mean that its:

Rsin(60) = 720.65
R = 832.13

So 832.13 would be the force in the y-component??
I only have one more guess left on my homework, so I want to make sure lol
 
  • #12
yb1013 said:
So would it go like 345.66cos(60) - 1020 = 0 ??

uhh? :confused:

perhaps I misunderstood your Rsin(theta) …

I assumed you meant that to be the y-component of the reaction force at the wall:

Ry = 720 + 200 + 80 - 345.66sin(60)

(ooh, and it's 1000 not 1020 :wink:)
 
  • #13
tiny-tim said:
uhh? :confused:

perhaps I misunderstood your Rsin(theta) …

I assumed you meant that to be the y-component of the reaction force at the wall:

Ry = 720 + 200 + 80 - 345.66sin(60)

(ooh, and it's 1000 not 1020 :wink:)

Yea forget what I just said, I had a brain fart for a second there lol

Its just 720, sorry about the confusion.

Thanks again!
 
  • #14
yb1013 said:
… Its just 720 …

better check that …
 
  • #15
well its right, I stated in one of my other posts that the numbers were exactly what my actual problem contained. the 720 N of the bear is actually 740, which is why when I added all the numbers up I got 1020 instead of 1000.. So in the end the force is equal to 720

Sorry for all the confusion
 
  • #16
stuffing down more goodies than the average bear …

yb1013 said:
the 720 N of the bear is actually 740, which is why when I added all the numbers up I got 1020 instead of 1000.

ah :rolleyes: … heavier than the average bear! :biggrin:​
 

Related to Find the tension in the wire and the components of the force exerted

1. What is tension in a wire?

Tension in a wire is the force that is applied to the wire in order to keep it taut and prevent it from sagging or breaking. It is also known as the pulling force exerted on the wire.

2. How is tension in a wire calculated?

Tension in a wire can be calculated by using the formula T = F * L, where T is the tension, F is the force applied to the wire, and L is the length of the wire. Alternatively, it can also be calculated by using the formula T = k * ΔL, where k is the spring constant of the wire and ΔL is the change in length of the wire.

3. What are the components of the force exerted on a wire?

The force exerted on a wire can be broken down into two components: the horizontal component and the vertical component. The horizontal component is the force that is parallel to the wire, and the vertical component is the force that is perpendicular to the wire. These components can be calculated using trigonometric functions.

4. How does the angle of the wire affect the tension and force exerted?

The angle of the wire can affect the tension and force exerted in two ways. Firstly, a smaller angle will result in a larger horizontal component of the force, leading to a higher tension in the wire. Secondly, a larger angle will result in a greater vertical component of the force, which can cause the wire to sag or even break if the force is too great.

5. Can the tension in a wire be greater than the force applied to it?

Yes, the tension in a wire can be greater than the force applied to it. This can happen when the wire is at an angle, as the tension will be equal to the force divided by the cosine of the angle. Therefore, even if the force applied is smaller, a smaller angle can result in a larger tension in the wire.

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