Question

A uniform 15m log of mass 150kg rests on two supports. One support is 3m from the left end, and the other is 4m from the right end. The forces exerted on the log by the supports are 

• F_{left end} =644 N, F_{right end} =828 N

• F_left end =828 N, F_{right end} =644 N.

• F_left end =736 N, F_{right end} =736 N.

• F_left end =644 N, F_{right end}=644 N.

• None of these answers is correct.

Answer 1

                     | <——— 15 m  —————-> |

                     |    3m   |                     |   4m    |

                     =======================  Weight / m = 98 N/m

                                 ^                    ^

                                  |          |         |

                                R1        V      R2

                                        1470 N

Weight per meter length , w = 150/15 = 10 kg = 10*9.8 = 98 N/m

R1 + R2 = 150 *9.8 = 1470 N

Taking moments about Left support :

1470* (7.5 - 3) - R2 * (15 - 4) - 98*3 *3/2  = 0

=> 11 R2 = 1470*4.5 - 98*9/2 

=> R2 = 6174/11 = 561 27 N (upward)

=> R2 = - 561.27 N (negative sign for upward direction)

=> R1 = 1470 - 561.27 = 908.73 N (upward)

=> R1 = - 908.27 N (negative sign for upward direction)

Option : None of these are correct : is the ANSWER.