A uniform board with a mass of 22 kg and a length of 8.5 meters rests horizontal

Question

A uniform board with a mass of 22 kg and a length of 8.5 meters rests horizontally on two supports. One support is 1.4 meters from the left end of the board. The amount of the normal force produced by the support on the left is 54 Newtons. Then, a person with a mass of 28 starts to walk towards the right end of the board. How far, in meters, from the right end of the board can he walk, before the board starts to come off the left support and rotate around the right support causing the person to fall off the board?

Correct answer: 0.75 (Please do not show the work if you do not get this answer) our professor also gave us this following explanation but I am sitll very lost

Explanation / Answer

Balancing torque about right support, let right support be at distance z from center of board.

F1*(L/2 - x+z) - Mg z = 0

54*(8.5/2 - 1.4 -z) = 22*9.8 z

z = 0.9523 m

Now, F1 will become zero when it will tip off,

Mgz = mgy

22*0.9523 = 28 y

y = 22*0.9523 /28

= 0.75 m answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.