Question

50 kg. Nz 12om/s...
or plz solve this also
$7kg \times n$
$1870 \times kg$
$36kg \times kg$
$2048 \times kg$
​

Answer 1

Answer:

55 N at K

1

and 43 N at K

2

Explanation:

Let the normal reactions from the knife edges (K

1

& K

2

) be N

1

,N

2

.

Now since the bar is in equilibrium,

N

1

+N

2

=mg

N

1

+N

2

=10×9.8N=98N

Also the system is in rotational equilibrium too.

Thus, about the center of the rod, the net torque must be zero.

Hence (6×9.8)(5)+(N

2

)(25)=(N

1

)(25)

⟹N

1

=55N

N

2

=43N

Hence normal reaction 55 N at K

1

and 43 N at K

2

are formed.