If both surfaces are smooth, then it is likely that the ladder would slide. If the floor is smooth and the wall is rough, then it is possible that the ladder would either stay stationary, or slide. The state (stationary/not stationary) would depend on if there is enough static friction provided by the wall. If the floor is rough and the wall is smooth, then it is possible that the ladder would either stay stationary, or slide. The state (stationary/not stationary) would depend on if there is enough static friction provided by the floor. The video above is for this case. If the floor is rough, and the wall is rough, then it is possible that the ladder would either stay stationary, or slide. The state (stationary/not stationary) would depend on if there is enough static friction provided by the floor and wall.
It doesn't make sense to talk about the 'normal force of the person' or the 'normal force of the ladder' because I am not taking into consideration the friction force between the person and the ladder. I have only analysed the person and the ladder as a combined system. The ground provides a force that is equal in magnitude but opposite in direction to the force the "ladder and person", as a combined system, provides to the ground. One way to think of part of this problem is, the ground provides a force. This force (provided by the ground) has a perpendicular component, and a parallel component. In this case, the downward force of the person, plus the downward force of the ladder, is equal to the upward force of the ground. The normal force is equal to the weight force of the ladder plus the weight force of the person in this case because the ground is horizontal. (see @3:48 to 4:20)
Ignore my comment above about "it doesn't make sense" as I was still trying to understand your comment. The normal force applied onto the person (if we drew a free body diagram of the person only) would be useful if I wanted to check if the person would slide, but we are not given information about the friction coefficient between the person and the ladder. The normal force applied onto the ladder by the ground (if we drew a free body diagram of the ladder only) would be the same as this free body diagram would also include a downward force due to the person. In this case, the combined system is sufficient to solve this problem. You may have a question which is a little more difficult than this one.
Hi all, there is a typo in the question. It should show a coefficient of 0.4 (Not 0.4 N)
What if the height is given
Thank you sir. Very informative.
Sir, why the normal force of the person and ladder are not considered?
Wow thank you have learnt
Sir can lean against smooth wall placed on a smooth floor?
If both surfaces are smooth, then it is likely that the ladder would slide.
If the floor is smooth and the wall is rough, then it is possible that the ladder would either stay stationary, or slide. The state (stationary/not stationary) would depend on if there is enough static friction provided by the wall.
If the floor is rough and the wall is smooth, then it is possible that the ladder would either stay stationary, or slide. The state (stationary/not stationary) would depend on if there is enough static friction provided by the floor. The video above is for this case.
If the floor is rough, and the wall is rough, then it is possible that the ladder would either stay stationary, or slide. The state (stationary/not stationary) would depend on if there is enough static friction provided by the floor and wall.
Cornelis Kok okk sir..thnq sir
Sir, why the normal force of the person and the ladder are not counted??
It doesn't make sense to talk about the 'normal force of the person' or the 'normal force of the ladder' because I am not taking into consideration the friction force between the person and the ladder. I have only analysed the person and the ladder as a combined system. The ground provides a force that is equal in magnitude but opposite in direction to the force the "ladder and person", as a combined system, provides to the ground.
One way to think of part of this problem is, the ground provides a force. This force (provided by the ground) has a perpendicular component, and a parallel component. In this case, the downward force of the person, plus the downward force of the ladder, is equal to the upward force of the ground.
The normal force is equal to the weight force of the ladder plus the weight force of the person in this case because the ground is horizontal. (see @3:48 to 4:20)
Ignore my comment above about "it doesn't make sense" as I was still trying to understand your comment.
The normal force applied onto the person (if we drew a free body diagram of the person only) would be useful if I wanted to check if the person would slide, but we are not given information about the friction coefficient between the person and the ladder.
The normal force applied onto the ladder by the ground (if we drew a free body diagram of the ladder only) would be the same as this free body diagram would also include a downward force due to the person.
In this case, the combined system is sufficient to solve this problem.
You may have a question which is a little more difficult than this one.
Sir what is the value of x`(xprime) plzz sir reply me
th-cam.com/video/N2mUvLtGMJY/w-d-xo.html
This is the next video which has the value for x'