Question

please give the correct answer with explanation.....and no spamming should be there or else your answer will be reported​

Answer 1

Answer:

In this type of questions , apply tge concept of Translational and Rotational equilibrium .

Let normal reaction at point A be N1 and that at point B be N2

As per Translational equilibrium , all the forces can be equated as follows :

$\sf{ \Sigma(forces) = 0}$

$\sf{= > N1 + N2 - 3 - 4 - 5 = 0}$

$\sf{= > N1 + N2 = 12 \: \: .....(1)}$

As per Rotational equilibrium , all the torques can be equated as follows :

Considering axis at point A ;

$\sf{ \Sigma(torques) = 0}$

$\sf{= > ( N2 \times 5) - (3 \times 2) }\\ \: \: \: \: \: \: \: \sf{ -( 4 \times 3) -( 5 \times 4)= 0}$

$\sf{= > 5( N2 ) = 38}$

$\sf{ \red{ \bold{= > N2 = 7.6 \: kN }}}$

Putting value of N2 in equation (1)

$\sf{= > N1 + 7.6 = 12 }$

$\sf{ \red{ \bold{= > N1 = 4.4 \: kN}}}$

Answer 2

$\underline{ \boxed{ \bold{ \rm{ \purple{ \huge{Answer}}}}}}$

Given :

A massless beam of length 5m is placed on two wedges A and B.

Three forces of 3kN,4kN and 5kN applied on the beam.

To Find :

Normal reaction at point A and B

Concept :

For translation equilibrium...

$\: \: \: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \bold{ \rm{ \pink{ \sum{(Forces)} = 0}}}}} \: \dag$

For rotational equilibrium...

$\: \: \: \: \: \: \: \: \dag \: \underline{ \boxed{ \bold{ \rm{ \blue{ \sum{(Torques)} = 0}}}}} \: \dag$

Calculation :

Let Normal reaction at point A be N1 and at point B be N2

$\red{ \star} \: \underline{\tt{Translation \: Equilibrium}} \\ \\ \implies \rm \: \sum{f} = N_1 + N_2 - 3 - 4 - 5 = 0 \\ \\ \therefore \rm \: \green{N_1 + N_2 = 12 \: kN} \\ \\ \red{ \star} \: \underline{ \tt{Rotational \: Equilibrium}} \\ \\ \implies \rm \: \sum{ \tau} = (5 \times N_2) - (2 \times 3) - (3 \times 4) \\ \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - (4 \times 5) = 0 \:\\ \\ \therefore \rm \: 5N_2 = 6 + 12 + 20 = 38 \\ \\ \therefore \: \underline {\boxed{ \bold{ \rm{ \orange{N_2 = 7.6 \: kN}}}}} \: \red{ \star} \\ \\ \implies \rm \: N_1 + N_2 = 12 \\ \\ \therefore \rm \: N_1 + 7.6 = 12 \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \purple{N_1 = 4.4 \: kN}}}}} \: \red{ \star}$