A seesaw consists of a board 4m long, which pivots about itscenter. A 50 kg pers

Question

A seesaw consists of a board 4m long, which pivots about itscenter. A 50 kg person sits at one end of the seesaw. How far fromthe pivot should a person of 75 kg sit that they can balancewithout touching the ground? A 1000 kg steel bar of length L is suspended at a point on thebar which is 0.5 m away from its center-of-mass. If a 50 kg personstanding at the end of the bar which is closer to the point ofsuspension is in rotational equilibrium, what is the total lengthof the bar? A seesaw consists of a board 4m long, which pivots about itscenter. A 50 kg person sits at one end of the seesaw. How far fromthe pivot should a person of 75 kg sit that they can balancewithout touching the ground? A 1000 kg steel bar of length L is suspended at a point on thebar which is 0.5 m away from its center-of-mass. If a 50 kg personstanding at the end of the bar which is closer to the point ofsuspension is in rotational equilibrium, what is the total lengthof the bar?

Explanation / Answer

If the board is 4m long, there are 2m on either side. The torquesfrom both of the people should be equal for the board to bebalanced. Therefore: (50kg)(9.8m/s2)(2m) =(75kg)(9.8m/s2)(d) d = 1.33m For the bar, given that the center of mass has to exist at thecenter of the bar, we can say that the center of the mass of thebar is at L/2. Again, the torque of the person must be equal thetorque of the bar. The distance of the person can be represented as.5m less than half of the length of the bar. (50kg)((L/2)-.5m)(9.8m/s2) =(1000kg)(9.8m/s2)(.5m) ((L/2)-.5) = 10m L/2 = 10.5m L = 21m

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