Question

A metal bar 70 cm long and 4 kg in mass supported on two knife - edges placed 10 cm from each end. A 6 kg weight 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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Explanation:

Figure shows the rod AB, the positions of the knife edges K1​ and K2​, the centre of gravity of the rod at G and the suspended load at P.

Note the weight of the rod W acts as its centre of gravity G. The rod is uniform in cross section and homogenous; hence G is at the centre of the rod;AB=70cm. AG=35cm, AP=30cm, PG=5cm, AK1​=BK2​=10cm and K1​G=K2​G=25cm. Also, W= weight of the rod =4.00kg and W1​= suspended load=6.00kg ; R1​ and R2​ 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 R2​ 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/s2.

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​.

Answer 2

Answer:

Figure shows the rod AB, the positions of the knife edges K

1

and K

2

, the centre of gravity of the rod at G and the suspended load at P.

Note the weight of the rod W acts as its centre of gravity G. The rod is uniform in cross section and homogenous; hence G is at the centre of the rod;AB=70cm. AG=35cm, AP=30cm, PG=5cm, AK

1

=BK

2

=10cm and K

1

G=K

2

G=25cm. Also, W= weight of the rod =4.00kg and W

1

= suspended load=6.00kg ; R

1

and R

2

are the normal reactions of the support at the knife edges.

For translational equilibrium of the rod, R

1

+R

2

−W

1

−W=0 ....(i)

Note W

1

and W act vertically down and R

1

and R

2

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

2

and W

1

are anticlockwise (+ve), whereas the moment of R

1

is clockwise (−ve).

For rotational equilibrium,

−R

1

(K

1

G)+W

1

(PG)+R

2

(K

2

G)=0 (ii)

It is given that W=4.00g N and W

1

=6.00g N, where g= acceleration due to gravity. We take g=9.8m/s

2

.

With numerical values inserted, from (i)

R

1

+R

2

−4.00g−6.00g=0

or R

1

+R

2

=10.00gN (iii)

=98.00N

From (ii),−0.25R

1

+0.05W

1

+0.25R

2

=0

or R

1

−R

2

=1.2g N=11.76N (iv)

From (iii) and (iv), R

1

=54.88N,

R

2

=43.12N

Thus the reactions of the support are about 55N at K

1

and 43N at K

2

.

Explanation:

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