A beam resting on two pivots has a length of l 600 m and...

A beam resting on two pivots has a length of L = 6.00 m and mass M = 82.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance ℓ = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 64.0 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip (b) Where is the woman when the normal force n1 is the greatest? c) What is n1 when the beam is about to tip? (d) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip. (e) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip. (f) Check the answer to part (e) by computing torques around the first pivot point.

Transcript

00:01 This would be our free body diagram or the sketch of the system essentially for part a.

00:05 And additionally, we have mg.

00:07 This would be equalling 90 .0 kilograms multiplied by 9 .80 meters per second squared.

00:17 And this is giving us 882 neutons.

00:20 We then have lowercase mg, the smaller mass of 55 kilograms, multiplied by 9 .80 meters per second squared, and this is giving us 539 neutins.

00:36 Now, at this point, we can say for part b, we know that when woman is at x equals 0, nub 1 is greatest, we know that woman is exerting max counterclockwise torque again at x equals zero so essentially n sub 1 will be exerting its maximum clockwise torque at that about the center in order to hold the beam in rotational equilibrium or static equilibrium so get this would be given that the torque is equaling zero so for part c then, we know that as the woman walks to the right along the beam, she'll eventually reach a point where the beam will actually start to rotate clockwise.

01:45 So her counterclockwise rotation diminishes, her counterclockwise torque diminishes as she walks right along the beam.

01:56 So here, once beam begins, to rotate, we can say the beam will start to lift off its left most.

02:28 We could say pivot point.

02:33 And essentially what that means is that the normal force will eventually decrease to zero.

02:37 So n -1 will equal 0 newtons.

02:43 For part d, we know that when the beam is right about the tip, the sum of the four forces in the y direction is equaling zero and the normal force sub one is equaling zero.

02:55 So this is essentially giving us that n sub two minus m g minus lowercase m g is equaling zero.

03:03 Therefore n sub two is equaling m g plus m g and this is giving us 882 newtons plus 539 neutins.

03:14 This is giving us 1 ,420 newtons approximately.

03:19 We have to round to three significant figures because we can't be more accurate than three significant figures.

03:28 For part e, then, we know that at the right most pivot, we can say that some of the torques at the right most pivot will equal zero.

03:43 And so at this point, we know that, again, n sub 1 is equaling 0...

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