Question 9

Author Message   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Monday, April 14, 2003 - 11:03 pm:    A uniform rod AB of length 1.8m and weight 40N rests on two smooth supports at C and D where AC = BD = 0.2m . A load is attached to A so that the supporting force at C is twice the supporting force at D. Find the magnitude of the load. Diameter, please post your solution here.   971165 Username: 971165 Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 02:33 am:    F=W/S as the supporting force at C = those at D W1/S1= 2*(W2/S2) S1=S2 W1= 20+y where y is the weight of the load W2= 20 therefore y = 20N the magnitude of the load = 2   971165 Username: 971165 Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 02:35 am:    haha I think it should be a wrong answer... but I really forget/don't know how to calculate it, sorry!   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Wednesday, April 16, 2003 - 12:28 pm:    So, you would better find a friend to help you, and post a certain solution.   Ming Username: 971004 Registered: 04-2003 Posted on Thursday, April 17, 2003 - 04:48 pm:    Let F be the force acted to the supports. take moment about the laod. 0.2x2F+1.6xF=40x0.9 =>F=18N then let f be the magnitude of the load. take moment tabout the c.g f x 0.9+0.7 x 18=0.7 x 36 f=14N right?   Lau Kai Kwong Username: lkk Registered: 04-2003 Posted on Thursday, April 17, 2003 - 04:59 pm:    Correct, but the description is unclear. You would better write : "Let F be the supporting force at D. Then the supporting force at C is 2F." "By taking moment about A, ..." Further, the load can be found in an easier way by equating the vertical forces as: W + 40 = 2F + F