i have a question, i have made at home a leg press set up which consists of an A frame bolted to the wall at the pointed end therefore acting as a hinge fixed at about waist height, the outer part of the A frame the wider part of the letter A i have loaded with weights at the moment about 85kg this weight is 30" inches from the wall with a timber prop holding it horizontal , I then lie on the floor with my feet up under the weight, i push the weight up vertically and remove the prop then lower the weight with my legs as far as possible then extend my legs to complete one rep. i repeat over and over until i stop and replace the prop. now, does the position of the hinge made a difference to the effort required? and because of the hinge am i actually lifting the full 85kg or less?
Hi, great video, im wondering about 15:37. You resolved the value of friction horizontally in part b, im confused why we also use this value (horizontal component of friction) as our vertical component of friction too, would appreciate the help, thanks!
When in limiting equilibrium, Friction = mu * reaction force at that point. 8mg/3 is the reaction force at the hinge. Therefore the friction (acting upwards in this case) is mu * 8mg/3. I am also a student so take my explanation with a pinch of salt if it's wrong :)
The above explanation is correct. There was never a frictional force acting horizontally. The 8/3 mg is the normal reaction force at the wall and can be found by resolving horizontally - so it is equal to the horizontal component of T. The vertical friction takes this and multiplies by mu
Hi, thanks for the excellent video. My question is, would the horizontal component of the force at D also be Tcos(theta).... same as that of the horizontal component at A?? Likewise, would the vertical component at D also be Tsin(theta), which is the same as Fr?? Cheers.
My best advice would be to try and ignore the forces at D. They are very different to what is happening at A and only a result of any tension force. Try to just see the problem as looking into the forces acting directly on the beam. If you are curious, essentially any force at D would be counteracting the force of tension pulling away from the wall. Think about a hook and string attached at a wall and pulling away from the wall with the string. There is a force at the wall pulling against tension to keep the hook in place - into the wall and up potentially.
A reaction force only exists when there is contact between objects. It is the result of one object imparting a force on another. At A the reaction force prevents the beam going through the wall for example. At B as there is no separate object preventing B from moving, there will be no reaction force.
Excellent content sir, keep it up! I learned all about hinges in mechanics.
What he said^
Beautifully explained. Thank you so much.
Amazingly well taught, clear as day. #Myteacher
i have a question, i have made at home a leg press set up which consists of an A frame bolted to the wall at the pointed end therefore acting as a hinge fixed at about waist height, the outer part of the A frame the wider part of the letter A i have loaded with weights at the moment about 85kg this weight is 30" inches from the wall with a timber prop holding it horizontal , I then lie on the floor with my feet up under the weight, i push the weight up vertically and remove the prop then lower the weight with my legs as far as possible then extend my legs to complete one rep. i repeat over and over until i stop and replace the prop. now, does the position of the hinge made a difference to the effort required? and because of the hinge am i actually lifting the full 85kg or less?
Watcbing this before a midterm i didnt study for
Lobstr get back to the sea earth is not a good place
Hi, great video, im wondering about 15:37. You resolved the value of friction horizontally in part b, im confused why we also use this value (horizontal component of friction) as our vertical component of friction too, would appreciate the help, thanks!
When in limiting equilibrium, Friction = mu * reaction force at that point. 8mg/3 is the reaction force at the hinge. Therefore the friction (acting upwards in this case) is mu * 8mg/3. I am also a student so take my explanation with a pinch of salt if it's wrong :)
The above explanation is correct. There was never a frictional force acting horizontally. The 8/3 mg is the normal reaction force at the wall and can be found by resolving horizontally - so it is equal to the horizontal component of T. The vertical friction takes this and multiplies by mu
@@huanggao4121 Haha no salt needed, thank you so much :)
@@westiesworkshop5983 Thank you so much!
Hi, thanks for the excellent video. My question is, would the horizontal component of the force at D also be Tcos(theta).... same as that of the horizontal component at A?? Likewise, would the vertical component at D also be Tsin(theta), which is the same as Fr?? Cheers.
My best advice would be to try and ignore the forces at D. They are very different to what is happening at A and only a result of any tension force. Try to just see the problem as looking into the forces acting directly on the beam.
If you are curious, essentially any force at D would be counteracting the force of tension pulling away from the wall. Think about a hook and string attached at a wall and pulling away from the wall with the string. There is a force at the wall pulling against tension to keep the hook in place - into the wall and up potentially.
@@westiesworkshop5983 Thank you, I appreciate your response.
Thank you 🙏
Why is there no reaction force at B?
A reaction force only exists when there is contact between objects. It is the result of one object imparting a force on another. At A the reaction force prevents the beam going through the wall for example. At B as there is no separate object preventing B from moving, there will be no reaction force.