Question

A body, weighing 10kg is suspended by two strings AC and BC at the point C .The length of the string AC=3m and BC =4m respectively. the horizontal distance of BC is 4 meters. find the tensions in the string AC and BC​

Answer 1

Answer:

Correct option is

C

5 kgf and 12 kgf

From the figure,

sinθ=

13

5

cosθ=

13

12

The horizontal force on the body balance each other.

Thus T

1

cosθ=T

2

cos(

2

π

−θ)

⟹T

1

cosθ=T

2

sinθ

Also the vertical components balance each other,

⟹13kgf=T

2

cosθ+T

1

sin(

2

π

−θ)

⟹13kgf=T

2

cosθ+T

1

sinθ

⟹T

1

=5kgf

T

2

=12kgf

Answer 2

Answer:

Since the system is in equilibrium;

The moment of forces about point A should be zero.

Thus;

10g×0.5+15g×1=Tsin53

o

×1

50+150=0.8T

T=250 N

At point A, the hinge force is working which can be resolved as R

y

& R

x

R

x

=Tcos 53

o

R

x

=250×0.6=150 N

R

y

=Tsin53

o

R

y

=250×0.8=200 N

R=

(R

x

)

2

+(R

y

)

2

R=

150

2

+200

2

R=250 N

Therefore, The tension in the string will be 250 N and the hinged force at point A will be 250 N