15.5) A uniform horizontal lever of length 6.0 m is pivoted at its mid-point. Th

Question

15.5) A uniform horizontal lever of length 6.0 m is pivoted at its mid-point. The following forces are acting on the lever: a 0.53 N vertically downward force acting at the right end of the lever, a 4 N vertically downward force acting at the left end of the lever, and a 5 N vertically downward force acting at a point on the lever 0.5 m away from the left end. Calculate the net torque acting on the lever.

The correct answer is 22.91 N m, but I need to know how to get there (formulas used, each step taken). Thank you!

Explanation / Answer

Solution:from the question we have

A uniform horizontal lever of length 6.0 m is pivoted at its mid-point.

Forces are acting on the lever: a 0.53 N vertically downward force acting at the right end of the lever, a 4 N vertically downward force acting at the left end of the lever, and a 5 N vertically downward force acting at a point on the lever 0.5 m away from the left end.

To find out-Calculate the net torque acting on the lever.

Let's consider counterclockwise as positive, and clockwise as negative forces

Then the pivot is at the centre, if you'd like draw the moment arm and the forces. Or just visualize them

Now, we have the left most side that is 3m away from the central pivot (since the rod is 6m) with a downward force of 4N. Torque is force x distance(from force application point to pivot)

o the Torque at the most left hand side is +12 (let's ignore units for now, since it's addition)

Now let's calculate the torque at the right hand most side: 0.53*3 but notice that is is counterclockwise. (I like to keep a pencil handy so I can use it as an aid for determining rotation)

since it is counterclockwise it will be negative so -0.53*3

The last part is 0.5 away from the left hand most side so this will be 2.5 away from the pivot. So given the formula it will be 2.5*5. Notice it is counterclockwise, so it is +2.5*5

To find the Net Torque we just add these "partial torques" and we get T= 2.5*5 + 4*3 - 0.53*3

Notice that we kept the signs of clockwise and counter clockwise since torque is a vector

Hence

Torque = 2.5*5 + 4*3 - 0.53*3=22.91 N m answer

Essentially in any circular motion vectors we ignore the cartesian coordiantes of x and y. Instead we describe the positives and negatives using clockwise and counterclockwise. How you assign the signs is arbitrary, as long as you note it at the beginning of the question

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