A uniform metal rod, with a mass of 3.0 kg and a length of 1.0 m, is attached to a wall by a hinge at its base. A horizontal wire bolted to the wall 0.64 m above the base of the rod holds the rod at an angle of 40 degrees above the horizontal. The wire is attached to the top of the rod. A) Find the tension in the wire.

Question

B)Find the horizontal component of the force exerted on the rod by the hinge.
C)Find the vertical component of the force exerted on the rod by the hinge.

Answer 1

Let me calculate each part one by one below.

A) To find the tension in the wire, we first have to find the torque acting on the rod due to gravity.

As we know the torque, τ = r × F, and F = mg

Here, r = 0.5 m (distance from the hinge to the rod's center of mass), m = 3.0 kg, g = 9.8 m/s^2

τ = 0.5 × 3.0 × 9.8 Nm = 14.7 Nm

Now, let's find the torque due to wire's tension. As τ = r × T × sin(θ)

Here, r = 1.0 m, θ = 40 degrees

So, 14.7 = 1.0 × T × sin(40)

T = 14.7 / sin(40) = 23.0 N (approx.)

B) Now, let's find the horizontal component of the force exerted on the rod by the hinge.

Since the rod is in equilibrium, the torque due to the horizontal force at the hinge must be equal to the torque due to the tension in the wire.

So, r × F_horizontal = τ_wire

F_horizontal = τ_wire / r = 14.7 / 0.64 = 22.9 N (approx.)

C) Finally, we need to find the vertical component of the force exerted on the rod by the hinge.

We know that the net force acting on the rod should be zero, as the rod is in equilibrium.

F_vertical - mg = 0

F_vertical = 3.0 × 9.8 N = 29.4 N

So, the answers are:

A) Tension in the wire: 23.0 N (approx.)

B) Horizontal component of the force exerted on the rod by the hinge: 22.9 N (approx.)

C) Vertical component of the force exerted on the rod by the hinge: 29.4 N