A metal bar 70 cm long
and 4.00 kg in mass supported on two
knife-edges placed 10 em from each end.
A 6.00 kg load is suspended at 30 cm from
one end. Find the reactions at the knife-
edges. (Assume the bar to be of uniform
cross section and homogeneous.)
Answers
Answer:
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Answer:
Explanation:
Figure shows the rod AB, the positions of the knife edges K1and K2 , the center of gravity of the rod at G and the suspended load at P.
AB=70cm
AG=35cm,
AP=30cm
PG=5cm
AK1 =BK 2 =10cm
and K1 G=K2 G=25cm
W= weight of the rod =4.00kg and W 1 = suspended load=6.00kg ; R 1 andR2 are the normal reactions of the support at the knife edges.
For translational equilibrium of the rod, R1 +R2 −W1 −W=0....(i)
Note W1 and W act vertically down and R1 and R2 act vertically up.
For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments of the forces is G. The moments of R and W1 are anticlockwise (+ve), whereas the moment of R1 is clockwise (−ve).
For rotational equilibrium,
−R1 (K1 G)+W1 (PG)+R2 (K2 G)=0 (ii)
It is given that W=4.00g N and W1 =6.00g N, where g= acceleration due to gravity. We take g=9.8m/s^2.
With numerical values inserted, from (i)
R1 +R2 −4.00g−6.00g=0
or R1 +R2 =10.00gN (iii)
=98.00N
From (ii),−0.25R1 +0.05W1 +0.25R2 =0
or R1 −R2 =1.2g N=11.76N (iv)
From (iii) and (iv), R1 =54.88N,
R2 =43.12N
Thus the reactions of the support are about 55N at K1 and 43N at K2.