Physics/Tension in Strings
QUESTION: Dear Sir,
I think you remember me. We're already in discussion about a numerical based on torque. I'll simply send an image in this message. Moreover I would like to clarify a point that I'm really in tenth grade and we've torque chapter. In all probability the difference in courses we've studied makes it sound strange. I would also like to ask if you'd mind giving me your email address, so that we can converse freely.
ANSWER: Hello AB,
About your earlier question: "A horizontal rod AB is suspended at its ends by two strings as given in the figure(image attached). The rod is 0.6m long and it's weight is 0.3kgf acting at G where AG is 40cm. Find the tension in the strings X and Y." Now that I have the figure, I can help you with it.
The rod is in equilibrium. Equilibrium means that it is not accelerating in linear motion or rotational motion. And that means that the net force, and the net torque, on the rod are both zero. To solve the problem, we will need equations showing both being zero. Tension will be involved in both equations. Since the force and torques are vector quantities, if we show them summing to zero, we would need to define positive and negative directions. I will instead show them on opposite sides of the equal sign so we are just dealing with magnitude. There are 2 tensions in this problem, both will be an upward force at either end of the rod. Let's call the tension in the left string T1 and use T2 for the tension in the right string.
Your concern when you sent this problem was your understanding of tension. The tensions in the strings provide an upward force on the rod. That is what we will be looking at here. But if you were concerned on the strength of the attachment of the screws in the ceiling at X and Y, you would need to understand that the string pulls in both directions. The downward force pulling at the screw at X has a magnitude of T1. (I do not typically work with the kgf unit. I typically would convert to Newtons. But I think I can adapt to kgf!)
Let's start with analyzing torque. We need to declare a pivot point. We will analyze potential for rotation about that point. This can be assigned just about anywhere on the rod. Some points have more advantages than others. I choose the left end, point A . There are 2 forces that would provide rotation about that pivot point. The weight of the rod acts as if it is all concentrated at point G and would tend to cause clockwise rotation about A. So it is a CW torque. The tension T2 would tend to cause counterclockwise (anti-clockwise) rotation about A. So it is a CCW torque. Since the rod is in equilibrium, these need to be equal and opposite.
CW torque = CCW torque
0.3 kgf*0.4 m = T2*0.6 m
T2 = 0.3 kgf*0.4 m / 0.6 m = 0.2 kgf
And now we analyze linear force to find T1. We already know T2 - so that is supporting some of the weight of the bar. Knowing the total weight of the bar, we can say
downward forces = upward forces
total weight = T1 + T2
0.3 kgf = T1 + 0.2 kgf
T1 = 0.1 kgf
I was glad that you have studied torque since this solution relied on those concepts. My earlier comments about you being in 10th grade were not expression of doubt, merely an expression of surprise that you are studying these difficult topics so early. I didn't until grade 12. Regarding my email address, I would be willing to send it to you if you select the "Private" option for your next question.
I hope this helps,
[an error occurred while processing this directive]---------- FOLLOW-UP ----------
QUESTION: Good morning Sir.
Thank you for your email address( I think follow ups are private questions). Please check out for an email I've sent you. In this question I'd like to ask you what if I use the other point say B and work out the tensions in the strings(the way opposite of what you did)?
Very sincerely yours,
Choosing B as the pivot point in the calculation should yield the same answers. I suggest that it would help you to try to adapt the method I used in my previous answer. If my explanation of what I was doing was clear enough, it should work for you. Good luck. But if not, let me know.
I hope this helps,