Answer :
Answer:
the horizontal force acting on rod is - 25.909 N
veritical force is 25.48 N
Explanation:
Given data:
mass of rod is m 2.6 kg
length of rod is 1.5 m
Hinge distance from the wire is 0.70 m
Apply torque law about hinge
[tex]T = (\frac{L}{2} cos 28^o) mg[/tex]
Tension in the wire is
[tex]T = \frac{[\frac{1.5}{2} cos 28 ]2.6 \times 9.8}{0.70}[/tex]
= -25.909 N
the horizontal force acting on rod is
Fx = T
FX = - 25.909 N
vertical force is
Fy = mg
[tex] = 2.6 \times 9.8 = 25.48 N[/tex]