Question

A homogenous board of 5.0 m in length and 260 N in weight is lied horizontally on two supports. The left support is located at the left end of the board and the right support is located at 1 m from the right end of the board. A person with a mass of 50 kg stand on the board at a distance of 0.9 m from the right end of the board. a) Draw the free body diagram of the forces that acted on the board. b) What is the magnitude of the forces exerted by each of the supports to the board? c) What is the distance (with respect to the right end of the board) the person can move so that the left end of the board starts to lift?

0,9 m If required, use gravity acceleration g = 10 m/s 1. A homogeneous board of 5.0 m in length and 260 N in weight is lied horizontally on two supports. The left support is located at the left end of the board and the right support is located at 1 m from the right end of the board. A person with a mass of 50 kg stand on the board at a distance of 0.9 m from the right end of the board. a. Draw the free body diagram of the forces that acted on the board. b. What is the magnitude of the forces exerted by each of the supports to the board? c. What is the distance (with respect to the right end of the board) the person can move so that the left end of the board starts to lift? in 1 m

Answer 1

9 FBD) - Om 0.9 m 2.5m - C. R2 sogn 26on b) Total force en verticle direction- Ry - 260+ 50x9.8 again moment about B 260x1.5 - Rix 4 - sox9. 80.1 -0 Ri=85.25N R 2 =-85.25 +750 - 664-75 N CJ - Distance som right and so that board starts to lift moment around B 260 X115-05-254 - 50x9.8*(1-x) so 49 < 490(1-x) 1-3 = 26% -0.9m for x<0.9 m Pt will stents to litaf.