Question

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.)​

Answer 1

Answer:

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Answer 2

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.