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

[yb1013]

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?

 

[Ashleyz]

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

 

[xxChrisxx]

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.

 

[yb1013]

o I am sorry that's what I forgot,

http://www.webassign.net/pse/p12-42.gif

thats the picture

 

[tiny-tim]

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.

 

[xxChrisxx]

o I am sorry that's what I forgot,

http://www.webassign.net/pse/p12-42.gif

thats the picture

Its yogi bear...!

Thats some nice pic-a-nic

 

[yb1013]

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

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

 

[Ashleyz]

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.

 

[tiny-tim]

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!

 

[yb1013]

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 ??

 

[yb1013]

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 want to make sure lol

 

[tiny-tim]

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:

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

(ooh, and it's 1000 not 1020 )

 

[yb1013]

uhh?

perhaps I misunderstood your Rsin(theta) …

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

R_y = 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]

… Its just 720 …

better check that …

 

[yb1013]

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]

stuffing down more goodies than the average bear …

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 … heavier than the average bear! ​