Question

Problem 3: Balance and Stability Consider the situation below, with a boy walking along a beam, which is resting on two supports (just resting on them – not physically attached). The beam is uniform with a mass of M = 70 kg and has a length of 5 m, and the boy has a mass of 45 kg. The supports are 3 m apart, and the boy is 0.4 m to the right of the right-hand support. The system is in static equilibrium and the beam is NOT in any danger of tipping over. A. Draw an extended free body diagram of the beam, and show all forces, including the forces from the left and right support. Label all forces clearly. B. Use the torque equation for static equilibrium to compute the force exerted on the beam from the left support. Hint: Pick your pivot point where the right support is located. C. Use the force equation for static equilibrium with your answer above to compute the force exerted on the beam from the right support.

          Problem 3: Balance and Stability
Consider the situation below, with a boy walking along a beam, which is resting on two supports (just resting on them – not physically attached). The beam is uniform with a mass of M = 70 kg and has a length of 5 m, and the boy has a mass of 45 kg. The supports are 3 m apart, and the boy is 0.4 m to the right of the right-hand support. The system is in static equilibrium and the beam is NOT in any danger of tipping over.

A. Draw an extended free body diagram of the beam, and show all forces, including the forces from the left and right support. Label all forces clearly.

B. Use the torque equation for static equilibrium to compute the force exerted on the beam from the left support. Hint: Pick your pivot point where the right support is located.

C. Use the force equation for static equilibrium with your answer above to compute the force exerted on the beam from the right support.

Show more…

Problem 3: Balance and Stability
Consider the situation below, with a boy walking along a beam, which is resting on two supports (just resting on them – not physically attached). The beam is uniform with a mass of M = 70 kg and has a length of 5 m, and the boy has a mass of 45 kg. The supports are 3 m apart, and the boy is 0.4 m to the right of the right-hand support. The system is in static equilibrium and the beam is NOT in any danger of tipping over.

A. Draw an extended free body diagram of the beam, and show all forces, including the forces from the left and right support. Label all forces clearly.

B. Use the torque equation for static equilibrium to compute the force exerted on the beam from the left support. Hint: Pick your pivot point where the right support is located.

C. Use the force equation for static equilibrium with your answer above to compute the force exerted on the beam from the right support.

Added by Harold B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Kirsty Gledhill

10/11/2023

Step 1

5 m from either end). - Weight of the boy (\(W_{boy} = mg\)) acting 0.4 m to the right of the right support. - Normal force from the left support (\(F_L\)). - Normal force from the right support (\(F_R\)). Show more…

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Transcript

00:01 Okay, so we've got this situation here and we want to start by drawing a free body diagram of all the forces present.

00:06 So we've got our beam and in the center of the beam its weight is going to act downwards.

00:14 So i'll call this w beam.

00:18 We've then also got the buoy and the weight of the buoy will also act downwards.

00:27 Then 0 .4 meters away from the buoy we have the reaction of the right hand support.

00:35 That's going to act upwards on the beam.

00:39 That's the weight of the beam pressing down on the support.

00:41 Newton's third law generates an equal and opposite force in response to that.

00:47 R2 and we've got the reaction force from the left hand support right the way over here which we'll call r1 and this is 2 .5 meters away from the center of the beam obviously and 3 meters away from the other support.

01:07 So that should be everything we need.

01:10 We then go on to part b which asks us to use the equilibrium of torque or of moments to find the reaction force at the left support which i've called r1 here.

01:26 So what we're doing is we're going to take the pivot to be our second.

01:31 So i shouldn't really put it there.

01:33 We'll take the pivot to be here at our second reaction.

01:40 So that's where we're measuring from.

01:43 So the forces so that's here.

01:47 The forces that are going clockwise then are r1 that's going clockwise and the weight of the buoy that's going clockwise as well...

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