Auniform rod is attached to a wall by a hinge at its base. the rod has a mass of 6.5 kg, a length of 1.3 m, is at an angle of 17° above the horizontal, and is held in place by a horizontal cord attached to the other end of the rod and bolted to the wall above the base of the rod.

(a) find the tension in the wire.
(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.

The tension in the wire is 104.18 N. The horizontal component of the force is 104.14 N. The vertical component of the force is 63.7 N

The tension in the wire refers to the net torque acting on the uniform rod and can be computed by using the formula:

Since the position of the rod is at rest, thus the torque acting on the rod = 0

∴

Making tension (T) the subject, we have:

Given that:

T = 104.18 N

Here, the horizontal component of the force exerted on the rod denotes the tensional force.

The vertical component of the force exerted on the rod is:

Learn more about tensional force here:

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tension wire 104 N, horizontal force hinge 104 N, vertical force hinge 63.7 N

Explanation:

The system described is in static equilibrium under the action of several forces, which are shown in the attached diagram, where T is the tension of the wire, W is the weight of the bar, Fx is the horizontal reaction of the wall and Fy It is the vertical reaction of the wall.

The directions of the forces are indicated by the arrows and are marked intuitively, but when solving the problem if one gives a negative value this indicates that the direction is wrong, but does not alter the results of the problem

For the resolution we use Newton's second law, both in translational and rotational equilibrium, if necessary.

We must establish a reference system to assign the positive meaning, we place it with the origin in the hinge, and the positive directions to the right and up. The location of the coordinate system allows us to eliminate the reaction of the hinge by having zero distance to origin. We write the equilibrium equations for each axis

     ∑ Fx = 0

     Fx -T =0

     ∑ Fy =0

     Fy -W =O      W = m g

     ∑τ =0

      -T dy + W dx =0

For rotational equilibrium we take the positive direction as counterclockwise rotation and distance is the perpendicular distance of the force to the axis of the coordinate system

       dy = L sin 17

      dx = (L/2) cos 17

Where L is the length of the bar and the weight is applied at the center of it, we write and simplify the equation

    -T L sin 17 + mg (L / 2) cos 17 = 0

   - T sin 17 + mg/2  Cos 17 =0

We write the three equations together

     Fx- T = 0

     Fy -W = 0

     -T sin 17 + (m g / 2) cos 17 = 0

a) With the third equation we can find the wire tension

     T = m g / 2 cos 17 / sin17

     T = 6.5  9.8/2 cotan 17

     T = 104 N

b) We use the first equation to find Fx

       Fx- T =0

       Fx = T = 104 N

c) We use the second equation to find Fy

       Fy = W = m g

       Fy = 6.5 9.8

       Fy = 63.7 N

massxgravity x heght=weightxheight= pe

4900=490xh

h=10

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