Physics Ninja shows how to solve for the contact forces between blocks sliding across a table. Check out my online class on Newton's laws of motion www.udemy.com/...
Explained very well but you could substitute the known variables earlier to make it simpler to understand. A lot of variables makes it confusing when you start learning the topic.
I agree, but it does help to get into the habit of solving for unknowns in variable form first, then plugging in all your known values into the final equation
@@max_feels Yeah I agree, its just that I thought for people trying to understand, substituting would be easier. But yeah agreed it helps to solve for unknowns first.
I used to think that way but it's far easier to understand conceptually when you solve everything algebraically and only add values at the end. Often times variables cancel out, and carrying tons of units throughout calculations becomes a nightmare and you end up confusing yourself. When I started only solving algebraically first, I started to truly grasp what exactly was going on.
very well explained, thank you very much. I tried to go to your webpage on udemy, but I could not find it. could you write the updated link please? Thanks
Substituting the values into the original equation to find both contact forces will give the same answer isnt? So, in that case, we take acceleration into consideration but in your equation even if do not acceleration we are able to find the forces so thank you for that information.
Vicente Martinez try adding the friction force on each free body diagram. You’ll still be left with 3 unknowns. (Acceleration, contact forces). Each block should have a friction force acting opposite of the motion
Explained very well but you could substitute the known variables earlier to make it simpler to understand. A lot of variables makes it confusing when you start learning the topic.
I agree, but it does help to get into the habit of solving for unknowns in variable form first, then plugging in all your known values into the final equation
@@max_feels Yeah I agree, its just that I thought for people trying to understand, substituting would be easier. But yeah agreed it helps to solve for unknowns first.
I used to think that way but it's far easier to understand conceptually when you solve everything algebraically and only add values at the end. Often times variables cancel out, and carrying tons of units throughout calculations becomes a nightmare and you end up confusing yourself. When I started only solving algebraically first, I started to truly grasp what exactly was going on.
FINALLY a video on this topic that makes sense. Thank you so much!!
Glad i was able to help!
Thank you so much man! I understood everything thanks again 😀
very well explained, thank you very much. I tried to go to your webpage on udemy, but I could not find it. could you write the updated link please? Thanks
Very good explanation thanks sir
phenomenal.
Thanks sir
I appreciate your help.
❤
So nice of you
Great! Could you make a video on "motion of bodies connected by a string"?
Thanks a ton!
I can't thank you enough...
Substituting the values into the original equation to find both contact forces will give the same answer isnt? So, in that case, we take acceleration into consideration but in your equation even if do not acceleration we are able to find the forces so thank you for that information.
thats great
Thanks sir
What would happen if there's friction
Vicente Martinez try adding the friction force on each free body diagram. You’ll still be left with 3 unknowns. (Acceleration, contact forces). Each block should have a friction force acting opposite of the motion
Online Physics Ninja physics ninja thank you very much
Vicente Martinez yes each block should have a different friction force pointing to the left. You can calculate the friction on each block.
Thankyouuuhhh😆
Thanks 👍👍👍👍😊😊😊
Good Example but the solution is unnecessarily complicated
Real concept
Bruhh that's 1st year physics ... we study dis in our school mn
Terrible, unnecessarily confusing
Reece Mason wank LOL. Happy learning