Calculate Tension of Wire with 6kg Mass on 3.5m Rod

[Chica1975]

A light rod of length 3.5m was pivoted at one end and was horizontal. On the end away from the pivot was a mass of 6kg. In addition to the pivot, a vertical wire supported the rod. The wire was attached at a point 1.16m from the pivoted end. Calculate the tension in the wire

Homework Equations

Force by gravity = mg
tan-1 = (opp/adj)
costheta = adj/hyp
sintheta = opp/hyp

The Attempt at a Solution

I drew a diagram relating to all the info given above. I have attempted to find the angle using mg and 3.5m i get something ridiculous like 1 degree. I know there are 180 degrees in a triangle and tried to find the other angle and follow this pattern using trigonometry until I was able to find the tension of the wire. My answer was way off.

apparently the correct answer is 177.

Thanks a mil.

 

[Lancelot59]

Can you create an image of the free body diagram you used? What are the actual calculations you did?

 

[Chica1975]

Unfortunately, I am traveling and do not have a scanner at all. However, I drew as best I could I also calculated mg = 58.8

I even tried to draw a right angled triangle using mg as one of the sides, this did not work.

The forces working on this are vertical (the wire), mg (gravitational).

Thanks

 

[Lancelot59]

Okay, I'll try and work you through the first bit of setting this problem up. From the looks if this the problem ignores the mass of the arm.

The thing with this problem is that it's not your straight up force problem, it requires you deal with torques. Never fear though, because they're incredibly simple to work with at this level. Just in case you've forgotten, all you need to calculate a torque is the force, and the distance from the pivot. So your torque is applied like this: Torque=(force in Newtons)(distance in metres), with your units being Newton metres (N*m).

Now as always with a free body diagram label all the forces acting on the object. I'm going to try and approximate the drawing here:

(pivot)====(support)=========(weight)

I'm going to define up, right, and counter-clockwise as positive. Imagine the support arrow going up, and the weight arrow going down.
Now when dealing with torques, a handy technique for eliminating unknowns is placing your pivot point on those unknowns. In this case you're given the pivot point, and the distances of each force applied from that pivot. So you don't need to worry about that.

Now use your standard net force equation, since the arm is stationary you have no net force. Therefore your torque forces must cancel out:
(the standard symbol for torques is a lowercase Tau)
$$F_{net}=0=\tau_{support}+\tau_{weight}$$

Try and move on from there.

 

[rwisz]

I would approach this problem by first identifying the forces acting on the system. In this case, we have the force of gravity acting on the mass (6kg) and the tension force in the wire. We can use Newton's Second Law (F=ma) to determine the magnitude of the tension force.

First, we need to find the acceleration of the mass. Since the rod is horizontal and the mass is not moving, we can assume that the acceleration is 0. Therefore, the net force acting on the mass must also be 0. This means that the tension force in the wire must be equal and opposite to the force of gravity acting on the mass.

We can calculate the force of gravity using the equation F=mg, where m is the mass (6kg) and g is the acceleration due to gravity (9.8 m/s^2). So the force of gravity on the mass is 6kg x 9.8 m/s^2 = 58.8 N.

Since the tension force in the wire is equal and opposite to the force of gravity, the magnitude of the tension force is also 58.8 N.

To find the angle of the wire, we can use trigonometry. Since the wire is attached at a point 1.16m from the pivoted end and the rod is 3.5m long, we can use the tangent function to find the angle. tan(theta) = 1.16m/3.5m = 0.3314. Taking the inverse tangent, we get theta = 18.4 degrees.

However, this angle does not match the given answer of 177 degrees. This may be a typo or mistake in the given information. If we assume that the wire is attached at a point 1.16m from the non-pivoted end instead, the angle would be 177 degrees. In this case, the tension force in the wire would be in the opposite direction, but the magnitude would still be the same (58.8 N).

In conclusion, the tension force in the wire is 58.8 N and the angle of the wire is either 18.4 degrees or 177 degrees, depending on the point of attachment.