Please verify if my answers are correct.

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The discussion revolves around a physics problem involving a beam, a bear, and a basket of food. The bear weighs 750 N, the beam weighs 200 N, and the basket weighs 80 N. The calculated tension in the wire when the bear is at 1.10 m is 372.1 N, with horizontal and vertical force components of 186.05 N and 707.75 N, respectively. The maximum distance the bear can walk before the wire breaks is determined to be 3.868 m.

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A hungry bear weighing 750 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 5.80 m long; the basket weighs 80.0 N.

Go to this link for the picture. http://answerboard.cramster.com/physics-topic-5-146405-0.aspx

(a) Draw a free-body diagram for the beam. (Do this on paper. Your instructor may ask you to turn in this sketch.)

(b) When the bear is at x = 1.10 m, find the tension in the wire and the components of the force exerted by the wall on the left end of the beam.

372.1 N (tension)
186.05 N ( Fx)
707.45 N ( Fy)

(c) If the wire can withstand a maximum tension of 750 N, what is the maximum distance the bear can walk before the wire breaks?

3.868 m


Part B) 0-(750)(1.10)-(200)(2.90)-(80)(5.80)+Tsin60(5.80)=0 so T=372.1 N

Fx=(372.1)cos60=186.05 N
Fy=1030-(372.1)sin60=707.75

Part C) T=750

(750)(sin60)(5.80)-(750)Xmax-(200)(2.90)-(80)(5.80)=0 solve for Xmax

Xmax=3.868 m

Are my answeres correct? I only have once chance to enter in my answer for webassign.

Thankyou
 
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What wire?

How is the wire placed in relation to the beam?
 
There is supposed to be a picture, but it won't generate on the website. Go to this link to see the picture. http://answerboard.cramster.com/physics-topic-5-146405-0.aspx
 
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