The *minimum* force required to keep the block in place is the downward force - the friction force trying to resist the downward force (25.2N - 7.5N = 17.7N). The *maximum* force that can be applied before the block starts sliding up is 25.2N + 7.5N = 32.7N.
Thank you for these videos , actually I have question regarding to the friction force, so why you applied the friction force in the same direction with the motion of the mass ?? As we know the mass is going to slide down and the friction force will resist this motion with the force F that is required in this problem !! am I right ?
The key word to this problem is equilibrium. If the problem would have asked for the minimum force to keep the block from sliding it would be 17.7 N, but it asked for equilibrium which implies there is zero net force. If only 17. 7 N is applied the friction force holding the block from sliding will still exist as the net force. (-7.5N)
“Friction always opposes motion” with that said... Wouldn’t the friction be pointing up instead since the block is moving downwards because of gravity?. Friction should “try” to prevent the object to fall down therefore the force applied to the object to create equilibrium should be less than the force applied by the selfweight? Or the other way around? Could you explain?, I am confused...
Your thought process is right, the only think is this questions asks what would the applied force be equal to for the block to stay in equilibrium, and since the force is applied upward, the friction force will oppose that motion. I hope this make sense and good luck!
I am still confused. I just feel that if I apply the force that you come up to the block it will move upward instead of being steady or in equilibrium. On the other hand, setting the friction upward, since we are not planning to create motion upwards just “stop” the body to continuing moving downward, we will come up with just enough force to set the body in equilibrium. If not would you please explain, I don’t want to screw this kind of questions in my FE. Thanks
@@rodolfovaldez4169 The thing is in this question, it clearly shows the direction of the force being applied. If you get a question where the direction of the force is not specified, that you go with that logic that you have. I hope this make sense and good luck with your studying!
I think the minimum force applied to prevent the block from sliding DOWN is 17.71N and the maximum force applied before the block begins to slide UP the ramp is 32.74N. Anything between these values and the block remains in equilibrium, and hopefully the exam is not tricky. So long as the coefficient of friction exceeds the tangent of the angle it will remain in equilibrium: u > tan 40 (without an external force)
At 5:02 in the video here I see the mg force going down and to the left how come it’s going that way and not down and to the right or straight down if you’re using the alternate interior angle of 40
The triangle that I drawn for the force mg just to show how we use to trigonometry functions to find the x and y component for the weight, but it should be going straight down.
Sorry but the answer in the video is wrong. The force F is compensating the friction difficiency to stop the body from moving downwards. F is most nearly 18 N.
I am solving a few exercises and I've seen a situation where they give you the weight already, so instead of kg, it's kN. The solution disregarded the angle and used the value in kN. I'm a little paranoic thinking that I might face a situation like that on the test. Is it unconventional?
If it is already in the force unit, it still needs to be broken down into x and y components, so you can apply the equilibrium equations. However, I have seen problems where they don't use the incline plane as x and y axis, they just use x and y axis to be straight, in that case the weight would just have the y component, but then the other forces need to be broken down into x and y components. So it depends how you choose your coordinate system. I hope this makes sense, and since you brought it up, I might make another video to clarify this issue.
I thought friction in this case is working against gravity. So doesn’t this add to the force?. Otherwise the applied force would break the friction which would result in the box moving.
Friction opposes motion, so if you apply a force upward to keep the object in equilibrium, the friction will oppose that motion. You can apply a force just enough to keep the object in equilibrium, however, if you keep increasing the force and the limiting friction is exceeded, that is when the object will start moving! I hope this makes sense and good luck with your studying!
enGENIEer ok thanks. I was thinking that gravity is naturally pushing on the box without an opposing force, causing it to slide.While friction was the only other force slowing it down. Meaning it was going in positive direction.
Hi Kenza, Thank you for sharing these informative videos. I watched all videos and followed your advice and tips. I passed my FE. Thanks a lot. If I wanna contact you for job related queries, is it possible to contact you and how?
I recalculated it three times and my results were 22.69 instead of 32.74. I got sin40=.745 and cos40=-0.6669 and when I did my calculations I got that result.
Hello, I am not really sure why you are getting 22.69, but if you plug in (4*9.81) [(0.25 cos 40) + (sin 40)], you should get 32.74. Just be careful when you plug in to your calculator, make sure you use the parenthesis correctly! So you should calculate this first, (0.25 cos 40) then add sin 40, which is 0.83 then multiply 0.83 by 4*9.81. I hope this helps and thank you for watching!
Hello Ropert, please note that the friction force has to always oppose your motion! So if you look at the problem carefully, the question is asking, what is the force that need to be applied upward to keep the system in equilibrium. So the force is going upward, so friction force has to go opposite of your force. I hope this makes sense. Thank you for watching and good luck with your studying!
@@ropertmores971 OK so that would be correct if the question did not show the force being applied upward. So think of it this way, if you apply that force, (assume it exceeds the limiting friction force) your block will start moving up, so hence your friction should be the opposite of that. Does that make sense?
@@Genieprep In this case we have two solutions to this force The first is that the force acting is only going to resist the weight and with the help of the friction force in this , The second force is greater than the first solution and therefore friction and weight will resist.
The *minimum* force required to keep the block in place is the downward force - the friction force trying to resist the downward force (25.2N - 7.5N = 17.7N). The *maximum* force that can be applied before the block starts sliding up is 25.2N + 7.5N = 32.7N.
You are absolutely right. Also, In the exam they will likely specify which one they want
Again a great run-down of this static problem...thank you
Thank you for watching and for your great support! I hope your studying is going well!
Et Deus Vult, with my blessings...Tomasi Akhengatan Tomasi Jr
Thank you for these videos , actually I have question regarding to the friction force, so why you applied the friction force in the same direction with the motion of the mass ?? As we know the mass is going to slide down and the friction force will resist this motion with the force F that is required in this problem !! am I right ?
The key word to this problem is equilibrium. If the problem would have asked for the minimum force to keep the block from sliding it would be 17.7 N, but it asked for equilibrium which implies there is zero net force. If only 17. 7 N is applied the friction force holding the block from sliding will still exist as the net force. (-7.5N)
“Friction always opposes motion” with that said... Wouldn’t the friction be pointing up instead since the block is moving downwards because of gravity?. Friction should “try” to prevent the object to fall down therefore the force applied to the object to create equilibrium should be less than the force applied by the selfweight? Or the other way around? Could you explain?, I am confused...
Your thought process is right, the only think is this questions asks what would the applied force be equal to for the block to stay in equilibrium, and since the force is applied upward, the friction force will oppose that motion. I hope this make sense and good luck!
I am still confused. I just feel that if I apply the force that you come up to the block it will move upward instead of being steady or in equilibrium. On the other hand, setting the friction upward, since we are not planning to create motion upwards just “stop” the body to continuing moving downward, we will come up with just enough force to set the body in equilibrium. If not would you please explain, I don’t want to screw this kind of questions in my FE. Thanks
@@rodolfovaldez4169 The thing is in this question, it clearly shows the direction of the force being applied. If you get a question where the direction of the force is not specified, that you go with that logic that you have. I hope this make sense and good luck with your studying!
I think the minimum force applied to prevent the block from sliding DOWN is 17.71N and the maximum force applied before the block begins to slide UP the ramp is 32.74N. Anything between these values and the block remains in equilibrium, and hopefully the exam is not tricky. So long as the coefficient of friction exceeds the tangent of the angle it will remain in equilibrium: u > tan 40 (without an external force)
Thank you
Hi, Thank you for sharing the video. mg is spited in to two mg cos and mg sin why mg sin is taken in negative direction?
At 5:02 in the video here I see the mg force going down and to the left how come it’s going that way and not down and to the right or straight down if you’re using the alternate interior angle of 40
The triangle that I drawn for the force mg just to show how we use to trigonometry functions to find the x and y component for the weight, but it should be going straight down.
So if we were to draw the forces straight down in the fbd it would still be the same way to find the c and y component
Either way thank you!!
Sorry but the answer in the video is wrong. The force F is compensating the friction difficiency to stop the body from moving downwards. F is most nearly 18 N.
I am solving a few exercises and I've seen a situation where they give you the weight already, so instead of kg, it's kN. The solution disregarded the angle and used the value in kN. I'm a little paranoic thinking that I might face a situation like that on the test. Is it unconventional?
If it is already in the force unit, it still needs to be broken down into x and y components, so you can apply the equilibrium equations. However, I have seen problems where they don't use the incline plane as x and y axis, they just use x and y axis to be straight, in that case the weight would just have the y component, but then the other forces need to be broken down into x and y components. So it depends how you choose your coordinate system. I hope this makes sense, and since you brought it up, I might make another video to clarify this issue.
@@Genieprep Thank you! It does make sense, but a video would definitely be helpful! Thank you!
I thought friction in this case is working against gravity. So doesn’t this add to the force?. Otherwise the applied force would break the friction which would result in the box moving.
Friction opposes motion, so if you apply a force upward to keep the object in equilibrium, the friction will oppose that motion. You can apply a force just enough to keep the object in equilibrium, however, if you keep increasing the force and the limiting friction is exceeded, that is when the object will start moving! I hope this makes sense and good luck with your studying!
enGENIEer ok thanks. I was thinking that gravity is naturally pushing on the box without an opposing force, causing it to slide.While friction was the only other force slowing it down. Meaning it was going in positive direction.
Hi Kenza, Thank you for sharing these informative videos. I watched all videos and followed your advice and tips. I passed my FE. Thanks a lot. If I wanna contact you for job related queries, is it possible to contact you and how?
Go to the ABOUT section of her youtube page.
@@matheussd93 Thank you
I recalculated it three times and my results were 22.69 instead of 32.74. I got sin40=.745 and cos40=-0.6669 and when I did my calculations I got that result.
Hello, I am not really sure why you are getting 22.69, but if you plug in (4*9.81) [(0.25 cos 40) + (sin 40)], you should get 32.74. Just be careful when you plug in to your calculator, make sure you use the parenthesis correctly! So you should calculate this first, (0.25 cos 40) then add sin 40, which is 0.83 then multiply 0.83 by 4*9.81. I hope this helps and thank you for watching!
@@Genieprep I found the problem, my calculator was set in rad instead of deg. So the trigonometric functions were wrong
Thank you so much for the support
@@SamequiPaiva I am glad to hear you figured it out and keep up the good work!
the direction of friction is wrong
Hello Ropert, please note that the friction force has to always oppose your motion! So if you look at the problem carefully, the question is asking, what is the force that need to be applied upward to keep the system in equilibrium. So the force is going upward, so friction force has to go opposite of your force. I hope this makes sense. Thank you for watching and good luck with your studying!
enGENIEer you mention that the fraction force is against the force “F” but it should be against the sliding direction of the block or body
@@ropertmores971 OK so that would be correct if the question did not show the force being applied upward. So think of it this way, if you apply that force, (assume it exceeds the limiting friction force) your block will start moving up, so hence your friction should be the opposite of that. Does that make sense?
@@Genieprep In this case we have two solutions to this force
The first is that the force acting is only going to resist the weight and with the help of the friction force in this
,
The second force is greater than the first solution and therefore friction and weight will resist.
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