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

## 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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is the beam supported by a horizontal string on each end?

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.

tiny-tim
Homework Helper
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!

I think you've calculated the y-component of the tension, instead of the y-component of the force at the wall.

would this be closer to the y-component for the force?

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

what a dumb bear...

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

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

tiny-tim
Homework Helper
Its yogi bear...!

Thats some nice pic-a-nic
He's smarter than the average bear!!
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!

what a dumb bear...

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

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

So would it go like 345.66cos(60) - 1020 = 0 ??

He's smarter than the average bear!!

Yup … if the tension is right, then that looks right too!
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 wanna make sure lol

tiny-tim
Homework Helper
So would it go like 345.66cos(60) - 1020 = 0 ??
uhh?

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 )

uhh?

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 )
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!

tiny-tim
Homework Helper
… Its just 720 …
better check that …

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

tiny-tim