Just have to say.. I've watched numerous videos on how to approach these kinds of problems. I've been stuck on one hw problem for the last 2 days and after watching your video I was able to solve it. Definitely going to be following your content and I'll be sure to pass the good word on to my friends at uni.
When a system is in static equilibrium, the net force on the system is zero (because there is no translational acceleration) and the net torque is zero (because there is no rotational acceleration). Net torque of zero means the clockwise and counterclockwise torques must have the same magnitude. z
Sir at the very end at the right hand side if we multiply 0.4 x 9.81 x 95 x 5 dont we get 1863.9 Nm and not 1426.4 as u have written in the equation. My second question is that to find the moment of the normal at the point where the surface is touching the wall we must take the perpendicular distance 5 sin 50 instead of just 5 in the above equation
1. I accidentally dropped a sin(theta) term (from the RHS of the first equation) from my writing, but I plugged it into the calculator correctly to get the right number for the result. 2. nw*sin(theta) gives you the perpendicular component of nw, then use the entire lever arm of 5 (again I accidentally dropped the sin(theta) after I made all the substitutions unfortunately), OR use the entire normal force multiplied by the perpendicular component of the lever arm. Either way you get the sin(theta) term. Run the numbers and you'll get my 1426N.
The assumption of the problem is that the wall is "smooth" in order to knock down the number of unknowns to a manageable level. Maybe not super realistic, but a decent approximation when you think about rubber feet on extension ladders vs. the smooth top on the side of a house. So we can call this a lower bound on the maximum height, or a decent "toy problem" anyway. z
I think the easiest way to handle it is to use the "translate the force vectors to the perpendicular lever arm" method. For example, mg for the person can be translated down to the floor, where the weight vector is perpendicular to your lever arm. The length of the perpendicular lever arm would then be Lcos(theta) - xcos(theta) for that horizontal distance to the rotation axis. Similar for the CM of the ladder. All other forces are already either perpendicular or parallel to the lever arm. For example, the torque exerted by the contact force with the wall is just n_w*Lsin(theta) in the CCW direction, where L*sin(theta) is the length of the lever arm. After that torque analysis, you can generate extra equations as needed by using the sum of forces in the x and y directions.
Zak. When you substituted Nw into your Torque balance and did your final calculation, what happened to the sin(theta) on the right side of the equation?
OMG, I can't believe I dropped a sine term without noticing -- that should be there! I'll add to the list of videos to redo this semester! New version will post late this month and I'll link in the description. -- Zak
I have seen in various other videos where the torque is calculated using sin theta. Is there a reason why cos theta is used here and if you use sin theta, will you be getting the same answer?
The important thing is that you take the component of force perpendicular to the lever arm. The way things are labeled here, that perpendicular component is given by a cosine (for the torque exerted by the weight of the ladder/weight of the person). This is because the angle between the vertical and the perpendicular is the same as the angle of incline for the ladder, and I made a quick "visual derivation" of that fact here: th-cam.com/users/shortsS8FnhFcoFxQ
Note there's an omission pointed out in the description: I left a factor of sin(theta) off the right hand side of the equation. However, I put the sin(theta) in my calculator when I was crunching the numbers! So the numbers are actually correct for the final answer here. z
I was looking for a tutorial to send to my niece. If you are looking for help and your physics teacher used sine to find lever arm, find a similar tutorial. This may confuse you more.
i tried to solve the same question as this. but the answer is wrong. it given coefficient of static friction between ladder and floor is 0.4 and the boy weighs is 500N. the length of the ladder is 2.2 m and the mass is 20kg. i got the answer for the height of the boy is 3.15, which is not logic because the length of the ladder only 2.2
I didn't have a chance to run your numbers, but using a lighter boy and shorter ladder can easily result in a ladder that never slips, no matter how high you go on it! It's very possible the math is just telling you "the ladder never slips". -- Zak
It's just that we are analyzing the net force and net torque on the ladder. The ladder must not move, so the net force on the ladder is zero and the net torque about any axis is also zero. The normal force on the man is not a force on the ladder, so it's not relevant to the analysis. -z
Just have to say.. I've watched numerous videos on how to approach these kinds of problems. I've been stuck on one hw problem for the last 2 days and after watching your video I was able to solve it. Definitely going to be following your content and I'll be sure to pass the good word on to my friends at uni.
High praise! Thanks -- Zak
Just curious what course were u doing in uni?
I have a problem, why is clockwise toque equal to anticlockwise toque in this case.
When a system is in static equilibrium, the net force on the system is zero (because there is no translational acceleration) and the net torque is zero (because there is no rotational acceleration). Net torque of zero means the clockwise and counterclockwise torques must have the same magnitude. z
Sir at the very end at the right hand side if we multiply 0.4 x 9.81 x 95 x 5 dont we get 1863.9 Nm and not 1426.4 as u have written in the equation. My second question is that to find the moment of the normal at the point where the surface is touching the wall we must take the perpendicular distance 5 sin 50 instead of just 5 in the above equation
1. I accidentally dropped a sin(theta) term (from the RHS of the first equation) from my writing, but I plugged it into the calculator correctly to get the right number for the result.
2. nw*sin(theta) gives you the perpendicular component of nw, then use the entire lever arm of 5 (again I accidentally dropped the sin(theta) after I made all the substitutions unfortunately), OR use the entire normal force multiplied by the perpendicular component of the lever arm. Either way you get the sin(theta) term. Run the numbers and you'll get my 1426N.
I have a problem, why the wall does not have a vertical upward reaction force (vertical upward friction) in this case? Thank you
The assumption of the problem is that the wall is "smooth" in order to knock down the number of unknowns to a manageable level. Maybe not super realistic, but a decent approximation when you think about rubber feet on extension ladders vs. the smooth top on the side of a house. So we can call this a lower bound on the maximum height, or a decent "toy problem" anyway. z
how would i approach this problem if i had to work out my calculations by taking the torques around the point where the wall meets the floor?
I think the easiest way to handle it is to use the "translate the force vectors to the perpendicular lever arm" method. For example, mg for the person can be translated down to the floor, where the weight vector is perpendicular to your lever arm. The length of the perpendicular lever arm would then be Lcos(theta) - xcos(theta) for that horizontal distance to the rotation axis. Similar for the CM of the ladder. All other forces are already either perpendicular or parallel to the lever arm. For example, the torque exerted by the contact force with the wall is just n_w*Lsin(theta) in the CCW direction, where L*sin(theta) is the length of the lever arm. After that torque analysis, you can generate extra equations as needed by using the sum of forces in the x and y directions.
Watching your video from India🇮🇳
cool! - Zak
Zak. When you substituted Nw into your Torque balance and did your final calculation, what happened to the sin(theta) on the right side of the equation?
OMG, I can't believe I dropped a sine term without noticing -- that should be there! I'll add to the list of videos to redo this semester! New version will post late this month and I'll link in the description. -- Zak
Also, I just discovered that I put in the sin(theta) when I ran the numbers, so the numerical result is correct in the video. Argh. - Zak
@@ZaksLab Thanks Zak. I didn't run through the calculation but I'm glad to know that you included it in yours.
I have seen in various other videos where the torque is calculated using sin theta. Is there a reason why cos theta is used here and if you use sin theta, will you be getting the same answer?
The important thing is that you take the component of force perpendicular to the lever arm. The way things are labeled here, that perpendicular component is given by a cosine (for the torque exerted by the weight of the ladder/weight of the person). This is because the angle between the vertical and the perpendicular is the same as the angle of incline for the ladder, and I made a quick "visual derivation" of that fact here: th-cam.com/users/shortsS8FnhFcoFxQ
How did you get 1426.4?
Note there's an omission pointed out in the description: I left a factor of sin(theta) off the right hand side of the equation. However, I put the sin(theta) in my calculator when I was crunching the numbers! So the numbers are actually correct for the final answer here. z
Enjoyed the video 💥
thanks! -- Zak
Thank you!
you're welcome! - Zak
I was looking for a tutorial to send to my niece. If you are looking for help and your physics teacher used sine to find lever arm, find a similar tutorial. This may confuse you more.
thank you
You're welcome! z
i tried to solve the same question as this. but the answer is wrong. it given coefficient of static friction between ladder and floor is 0.4 and the boy weighs is 500N. the length of the ladder is 2.2 m and the mass is 20kg. i got the answer for the height of the boy is 3.15, which is not logic because the length of the ladder only 2.2
I didn't have a chance to run your numbers, but using a lighter boy and shorter ladder can easily result in a ladder that never slips, no matter how high you go on it! It's very possible the math is just telling you "the ladder never slips". -- Zak
Can u please explain Why do we not consider the normal reaction exerted on the man by the ladder?
It's just that we are analyzing the net force and net torque on the ladder. The ladder must not move, so the net force on the ladder is zero and the net torque about any axis is also zero. The normal force on the man is not a force on the ladder, so it's not relevant to the analysis. -z
@@ZaksLab Oh I understand now. Thank you very much for the clarification 👍
This is my power! Full idol power!
nice!
Thanks! I can't wait to get back to it next week!
So easy with lagrange
Some day I'll do a lagrangian mechanics playlist . . . z