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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 [Broken]

(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

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

(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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