A large uniform \'butcher block\' of mass 100 kg rests on two supports and has a
ID: 1980754 • Letter: A
Question
A large uniform 'butcher block' of mass 100 kg rests on two supports and has a weight hanging from its end. The block has a mass of 100 kg and a length of 2 meters, if L=1.68 meters and the hanging weight is 178 Newtons what is the force on the left support to the nearest newton? The answer is 363 In the previous problem what is the force on the right support? The answer is 796 In problem 6 if the hanging weight is 126 newtons what is the minimum value for L for the configuration to remain stable to the nearest hundredth of a meter? The answer is 1.11 In problem 6 what is the minimum value for the hanging weight to the nearest tenth of a kilogram if L=1.37 meters and the configuration is to remain stable?Explanation / Answer
The principle behind this is simple = for a body to be balanced both the net torque
and force on the body should be zero
let the force on the left support be n1 and that of right support = n2
torques about the right support must be zero
so
n1*1.68-100*9.8*0.68+178*(2-1.68)=0
hence n1=362.7619 Newton
since n1+n2=100*9.8+178 (balance of forces)
n2=795.238 Newtons
let the system be on the verge of tipping for the minimum l
note the system tips only about right support
on the verge of tipping n1=0;
the net torques about rght support are balanced hence
100*(L-1)*9.8=126*(2-L)
1106*L=980+126*2
L=1.1139metre
similar to the case above
100*9.8*(1.37-1)=w*(2-1.37)
W=575.5556 newton...:)
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