A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is : a)b)c)d)mgCorrect answer is option 'B'. Can you explain this answer?

Question

A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is :       a)b)c)d)mgCorrect answer is option 'B'. Can you explain this answer?

Accepted Answer

The distance of Centre Of Mass of the given right angled triangle is 2L/3 along BA and L/3 along AC from the point B.
Force of magnitude mg is acting downwards at its COM.
Moment balance around B gives:
mg(2L/3)−F_A(L)=0
(Moment= × =rFsin(θ)=F(rsin(θ))=Fr⊥)
∴F_A=2mg/3

Answer

On fixing A as origin the co-ordinates of other vertices are B(L,0) and C(0,-L). Now, weight of any body acts at center of mass and for a triangle center of mass will act at the centroid. The coordinates of centroid G is (L/3,-L/3).Since we don't know in which direction reaction force acts, we will assume reaction force has two components Rx and Ry. Since the triangle is in equilibrium net horizontal forces, net vertical forces and net torque will all be zero.

Answer

as

the

triangle

is

in

equilibrium

:

Let

us

assume

the

horizontal

component

of

the

reaction

force

to

be

H

and

the

vertical

component

of

the

reaction

force

at

the

hinge

A

to

be

V

let

the

normal

force

acting

on

the

support

at

B

=

N

the

mass

of

the

plate

is

m

so

,

for

translational

equilibrium

along

the

horizontal

axis

:

H

=

0

while

for

the

vertical

equilibrium

:

V

+

N

−

mg

=

0

so

V

+

N

=

mg

for

rotational

equilibrium

:

taking

torque

of

all

the

forces

about

the

point

B

we

have

,

mg

(

2

l

/

3

)

=

Vl

or

V

=

2

mg

/

3

while

H

=

0

hence

the

reaction

force

on

the

plate

is

2

mg

/

3

i

n

t

h

e

v

e

r

t

i

c

a

l

y

u

p

w

a

r

d

i

r

e

c

t

i

o

n

Answer

1st make a equation balancing weight mg with the reaction from attached point A and the string.

mg = T+N(n being normal from A)

2nd make a rotation equilibrium equation, balancing torque from string and from A, taking centre of mass of triangle for balancing.
terminate the value of T from the rotational equation to get desired value of N

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