Physics Help Forum Box w/ horizontal force up incline w/ friction

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 Aug 3rd 2018, 03:26 PM #1 Junior Member   Join Date: Aug 2018 Posts: 5 Box w/ horizontal force up incline w/ friction Alright, I have spent the last 9 hours trying to figure out how on earth I am supposed to determine the horizontal force in this problem. https://drive.google.com/file/d/1dtu...ew?usp=sharing I've got to be missing something obvious. I know that the answer to (a) is supposed to be 106, and once I have that I can easily figure out everything else. I just have no idea where that number came from! Any help is appreciated! *If the link doesn't work, here's the description of the problem. Block M = 7.50 kg is initially moving up the incline and is increasing speed with a=4.03 m/s^2. The applied force F is horizontal. The coefficients of friction between the block and incline are fs=0.443 and fk=0.312. The angle of the incline is 25.0 degrees.
 Aug 4th 2018, 06:36 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 So where does your problem lie? Can you not write down a Newton's second law equation (do you know what this means?) relating the (known) acceleration of the block up the (known) slope to the forces acting? There is only one unknown in that equation so you should then be able to solve it for your answer. Edit (and hint) Is there another part to this question since they have given you both the static and dynamic coefficients of friction? Last edited by studiot; Aug 4th 2018 at 06:38 AM.
 Aug 4th 2018, 10:25 AM #3 Junior Member   Join Date: Aug 2018 Posts: 5 The only second law equation I know is F=ma, which gives me the net force. Is there a way to calculate it with the acceleration and slope? I haven't seen anything like this question in any of the textbook examples. I'm taking an online course (bad decision) and basically have to teach myself out of the textbook. Does it have something to do with calculating the amount of force needed to move the block with the static friction coefficient? I've only learned how to calculate friction forces with normal force which I don't know how to get without the horizontal component of the push force.
Aug 4th 2018, 01:28 PM   #4
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 Originally Posted by Christianna The only second law equation I know is F=ma, which gives me the net force. Is there a way to calculate it with the acceleration and slope? I haven't seen anything like this question in any of the textbook examples. I'm taking an online course (bad decision) and basically have to teach myself out of the textbook. Does it have something to do with calculating the amount of force needed to move the block with the static friction coefficient? I've only learned how to calculate friction forces with normal force which I don't know how to get without the horizontal component of the push force.
Don't worry, there's plenty of help for those with a serious interest in the subject.

Yes Newton's second law (N2) is Force = mass times acceleration

I assume you have covered resolution of forces so you know how to resolve forces parallel to and perpendicular to the slope.

A pity you have used F for the horizontal force, not H.
So I will use Z for the frictional force, and R for the reaction.

The block is in equilibrium perpendicular to the slope (or if you like has zero acceleration perpendicular to the slope) so resolving perp to the slope

R = Zsin25 + Mgcos25 (or M times zero = Zsin25 + Mgcos25 - R)

The block is moving and not in equilibrium parallel to the slope so applying N2

Ma = Fcos25 -Mgsin25 - Z

Finally we have the relation that Z = the coefficient of dynamic friction x R.

Putting these three equations together allows us to solve for F.
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Last edited by studiot; Aug 4th 2018 at 03:20 PM.

 Aug 4th 2018, 05:23 PM #5 Junior Member   Join Date: Aug 2018 Posts: 5 Thank you so much! I kept forgetting that the perpendicular acceleration would be zero since the block obviously isn't floating away (duh!). I knew I was missing something obvious.
Aug 4th 2018, 05:29 PM   #6

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 Originally Posted by Christianna Thank you so much! I kept forgetting that the perpendicular acceleration would be zero since the block obviously isn't floating away (duh!). I knew I was missing something obvious.
So obvious that practically all students that took my PHYS I course in college made it at some point or another. Don't worry, you are in good company.

-Dan
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