Question

A uniform metal rod, with a mass of 2.6 kg and a length of 1.5 m, is attached to a wall by a hinge at its base. A horizontal wire bolted to the wall 0.70 m above the base of the rod holds the rod at an angle of 28 ∘ above the horizontal. The wire is attached to the top of the rod.Find the tension in the wire.Find the horizontal component of the force exerted on the rod by the hinge.Find the vertical component of the force exerted on the rod by the hinge.

Answer 1

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

$T = (\frac{L}{2} cos 28^o) mg$

Tension in the wire is

$T = \frac{[\frac{1.5}{2} cos 28 ]2.6 \times 9.8}{0.70}$

   = -25.909 N

the horizontal force acting on rod is

Fx =  T

FX = - 25.909 N

vertical force is

Fy = mg

    $= 2.6 \times 9.8 = 25.48 N$