Hey mates plz solve this........no spam

1.	Hey mates plz solve this........no spam
Answer» EXPLANATION: consider this to be a case of a simply supported BEAM AB. the supports being at both the end points and let the length of the beam be 1.8m. assume the weight of the CAR to be a point load acting at a distance of 1.05m from poi t B. applying the laws of equilibrium, summation of all forces in the vertical direction gives F(A) + F(B)=1800...1 also moment balance at point A gives 18000.75=F(B)1.8....2 F(B)= 750Kg so F(A)=1050Kg the load on each front TYRE is 750/2=375kg the load on each back tyre is 1050/2=525kg. hence solved 2700=1.8B B= 1500kg A= 300kg

Answer»

consider this to be a case of a simply supported BEAM AB. the supports being at both the end points and let the length of the beam be 1.8m.

assume the weight of the CAR to be a point load acting at a distance of 1.05m from poi t B.

applying the laws of equilibrium,

summation of all forces in the vertical direction gives F(A) + F(B)=1800...1

also moment balance at point A gives

1800*0.75=F(B)*1.8....2

F(B)= 750Kg

F(A)=1050Kg

the load on each front TYRE is 750/2=375kg

the load on each back tyre is 1050/2=525kg.

hence solved

2700=1.8B

B= 1500kg

A= 300kg

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