Question

A uniform horizontal rod weighing 98 N is pivoted at a vertical wall. Its far
end is supported by a string whose other end is fixed to a point on a wall above
the pivot. A boy of mass 20 kg stands at the center of the rod. Find only the
vertical component of the tens ion in the string and only vertical component of
the reaction at the pivot. Length of the horizontal rod is 4 m. ​

Answer 1

Answer:

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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