Physics book, practice passage pg 52, Qs 24
Physics book, practice passage pg 52, Qs 24
I have been struggling with these questions because I am not understanding where the equation to calculate tension is coming from. For Q2, I understand that the clockwise and counterclockwise torque will be equal to each other at equilibrium, but then I do not know why the counterclockwise torque is set up as "10 = T(0.25)" which is then stated as the vertical component of the tension in the string. The solved answer then goes on to say that "tension in the string is T = 40/sin 60". Again, I do not understand where these tension equations come from, or how they are derived, so I am struggling to solve these problems on my own. For Q3, I also do not understand how tension in the x direction is calculated because I have not seen tension equations like these anywhere in the chapter. The answer solution to Q4 also seems to be off because it provides the solution to the wrong experiment. Bottom line, if I could learn where tension came from, I would feel much better about these problems. Thanks.

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Re: Physics book, practice passage pg 52, Qs 24
Thanks for your question!
Remember that for torque, we are only interested in the forces that are perpendicular to the beam. This is because anything horizontal will be cancelled out by an opposing force from the hinge.
The tension force is coming from the “massless string” that’s attached to the beam and to M1. So, really, the tension force is just the force of gravity on M1. Since M1 isn’t moving, the tension force must be opposing the only other force (Fg) acting on the block. So, Fg = M1 * g (which is equal but opposite the force of tension). From there, you can break the Ft down based on the angle (theta = 57° or roughly 60°).
Hopefully that helps clear up the tension force and the rest of it should be clearer. Happy to help further if you have more questions!
Remember that for torque, we are only interested in the forces that are perpendicular to the beam. This is because anything horizontal will be cancelled out by an opposing force from the hinge.
The tension force is coming from the “massless string” that’s attached to the beam and to M1. So, really, the tension force is just the force of gravity on M1. Since M1 isn’t moving, the tension force must be opposing the only other force (Fg) acting on the block. So, Fg = M1 * g (which is equal but opposite the force of tension). From there, you can break the Ft down based on the angle (theta = 57° or roughly 60°).
Hopefully that helps clear up the tension force and the rest of it should be clearer. Happy to help further if you have more questions!
Re: Physics book, practice passage pg 52, Qs 24
Thanks for your help! I also have another question regarding the Physics book, Chapter 1 practice passage, #3, pg. 22. I understand how to eliminate both B and D because of the 45 degree angle, but I am not following the explanation for the answer too well. I do not get how it is clearly determined that it is the angle with the vertical and NOT the angle with the ground that is likely to cause the Colles fracture. Could you please explain how this is determined?

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 Joined: Fri May 25, 2018 9:15 am
Re: Physics book, practice passage pg 52, Qs 24
Well we're basically trying to find the force along the arm and aren't actually interested necessarily in the angle. You could solve this question with respect to either angle and then solve for the third one as needed.
We know that the man's weight is 600N and we know a force at or greater than 1150 N would result in a broken arm. So we can basically set the hypotenuse as the arm and call it 1150 (or for the sake of math ease, 1200N!). Now we're looking at a right angle triangle with two known sides and we can solve for one of the other two angles. (The other known side is the vertical one, the man's weight, 600N.) Keep in mind that solving for one angle will give us the other, since we know that the three angles must add up to 180°.
Now you can solve for the angle with respect to the ground or the vertical, it doesn't actually matter. What you'll end up determining is that the angle needs to be 60 with respect to the vertical (an option in the answer choices) and 30 with respect to the ground (not an option in the answer choices). If you solved for the 30 with respect to the ground and that's not an option, solve for the unknown (or vertical) angle: 180°  90°  30° = 60° = angle with respect to the vertical.
To rephrase and summarize, we know the man's weight is the vertical side of the triangle and the arm is the hypotenuse. From there, you can solve any angle or side you'd like using the rules of trigonometry.
Hope this helps!
We know that the man's weight is 600N and we know a force at or greater than 1150 N would result in a broken arm. So we can basically set the hypotenuse as the arm and call it 1150 (or for the sake of math ease, 1200N!). Now we're looking at a right angle triangle with two known sides and we can solve for one of the other two angles. (The other known side is the vertical one, the man's weight, 600N.) Keep in mind that solving for one angle will give us the other, since we know that the three angles must add up to 180°.
Now you can solve for the angle with respect to the ground or the vertical, it doesn't actually matter. What you'll end up determining is that the angle needs to be 60 with respect to the vertical (an option in the answer choices) and 30 with respect to the ground (not an option in the answer choices). If you solved for the 30 with respect to the ground and that's not an option, solve for the unknown (or vertical) angle: 180°  90°  30° = 60° = angle with respect to the vertical.
To rephrase and summarize, we know the man's weight is the vertical side of the triangle and the arm is the hypotenuse. From there, you can solve any angle or side you'd like using the rules of trigonometry.
Hope this helps!
Re: Physics book, practice passage pg 52, Qs 24
Thank you, that makes sense! However, I am still greatly struggling with Q2 from pg. 53. I have most everything figured out until the last portion when it calculates the Ft of the massless string that goes from the wheel to the beam. I figured out how the torque of the vertical component of the string is equal to 40 N, but I do not get the part of the solution that sets the Ft of the xcomponent equal to torque divided by sin theta " T = (40) / (sin 60)" When I go to solve this, I end up getting this "T = (40) / (d)(sin 60)" which would equals 100 N instead of the book answer of 40 N. I am not sure why the book solution disregards the distance when calculating this problem. Could you please explain the following.
1. Why is distance disregarded when calculating the tension force in the xcomponent of the string?
2. Why does the xcomponent of the string use the 60 degree angle as theta, but all of the other torque calculations use the 90 degree angle as theta?
1. Why is distance disregarded when calculating the tension force in the xcomponent of the string?
2. Why does the xcomponent of the string use the 60 degree angle as theta, but all of the other torque calculations use the 90 degree angle as theta?

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Re: Physics book, practice passage pg 52, Qs 24
Distance is relevant when calculating torque, but it's distance from the center of rotation that we're interested in. In this problem, there is no need to calculate the xcomponent of the string, because, with torque problems, it's only the ycomponent of any force that we'd be interested in. (The xcomponents is cancelled out by the hinge.) In calculating tension forces, there is no distance involved.
The ycomponent of the tension force from the string is calculated using sin60° because the angle between the string and the beam is 60°. For the other two torques, we have the beam which is all vertical component (so the angle is 90° and sin90°=0) and the block M2 which is also all vertical component (so the angle is 90° again). The string doesn't exert its force at a 90° angle and so we must calculate what component of the force is at a 90° angle. This will be the ycomponent. The distance is included in the torque  the string is attached about midway down the beam (you can find this using geometry rules). 1/2 of the beam is 0.25, which you will see included in the explanation.
Ft of the string will be M1 * g. We need to find M1 such that the vertical component of the tension force will be equal in magnitude to the clockwise torques calculated before.
The ycomponent of the tension force from the string is calculated using sin60° because the angle between the string and the beam is 60°. For the other two torques, we have the beam which is all vertical component (so the angle is 90° and sin90°=0) and the block M2 which is also all vertical component (so the angle is 90° again). The string doesn't exert its force at a 90° angle and so we must calculate what component of the force is at a 90° angle. This will be the ycomponent. The distance is included in the torque  the string is attached about midway down the beam (you can find this using geometry rules). 1/2 of the beam is 0.25, which you will see included in the explanation.
Ft of the string will be M1 * g. We need to find M1 such that the vertical component of the tension force will be equal in magnitude to the clockwise torques calculated before.
Re: Physics book, practice passage pg 52, Qs 24
Hello,
I am still not quite grasping question 2 on practice passage 2. I understand how to calculate the clockwise torque. I am unsure of what is meant by 10=T(0.25), or where this was derived from. I understand that the torques must cancel out since they are at equilibrium. I have looked at all the forums regarding this question and have not received the clarity i hoped for. Appreciate any help. Thank you
I am still not quite grasping question 2 on practice passage 2. I understand how to calculate the clockwise torque. I am unsure of what is meant by 10=T(0.25), or where this was derived from. I understand that the torques must cancel out since they are at equilibrium. I have looked at all the forums regarding this question and have not received the clarity i hoped for. Appreciate any help. Thank you

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Re: Physics book, practice passage pg 52, Qs 24
This is essentially saying that the vertical component of tension is going to generate torque via this lever. We know therefore that the vertical component of tension has to generate CCW torque equal to the CW torque. So what we do is we first solve for the vertical component of tension, then use trigonometry to solve for the total tension in the string (knowing that this has to be equal to the weight [force felt by the mass] of m1).
I feel these questions are usually not that well answered in pure text, so I've attached a solution pathway. Let me know if that helps!
I feel these questions are usually not that well answered in pure text, so I've attached a solution pathway. Let me know if that helps!
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Re: Physics book, practice passage pg 52, Qs 24
This is exactly what I needed. Thank you so much!