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Newton's Laws question

  • Thread starter corong1997
  • Start date
  • #1

Homework Statement


A 880 N crate is being pulled across a level floor by a force F of 385 N at an angle of 27° above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.25. Find the magnitude of the acceleration of the crate.


Homework Equations


Fnet=ma


The Attempt at a Solution


Okay, so I drew out a diagram. You need to find Force Normal in the X direction in order to get FF=coefficientFN, so the equation for the X direction would be FAcosTheta+FN-mg=ma. That eventually turns into FAcosTheta-mg=FN. The problem is, once I plug all the values into the Y equation, the website says I'm wrong. Please help me!
 

Answers and Replies

  • #2
Simon Bridge
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Okay, so I drew out a diagram.
Good to see.
You need to find Force Normal in the X direction in order to get FF=coefficientFN, so the equation for the X direction would be FAcosTheta+FN-mg=ma. That eventually turns into FAcosTheta-mg=FN.
You've mixed horizontal and vertical forces in one expression.

On your diagram, draw out the components of F.
[itex]F\cos\theta[/itex] would be the horizontal component of the force - FN ([itex]F_N[/itex]) would be vertical right?
 
  • #3
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F = MA
you know the box is 880N , so divide that by gravity to get the mass.
The acceleration is in the horizontal, so you would use the horizontal force.
make a right triangle with 385N as the hypotenuse, and do COS(27) *385 to get the horizontal force.
But, you need the net force. so to find FF, you'd do mew*normalforce
the normal force, is the net vertical forces. So, you'd need to do 880N - the vertically applied force component (Sin27)*385
so you'd do .25 * (Sin27)*385

now, Fh - FF to get the net horizontal force

A = F / M

I got 1.857 m /s
 
  • #4
Simon Bridge
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@PotentialE - not bad, just a few pointers:
F = MA
you know the box is 880N , so divide that by gravity to get the mass.
The acceleration is in the horizontal, so you would use the horizontal force.
make a right triangle with 385N as the hypotenuse, and do COS(27) *385 to get the horizontal force.
But, you need the net force. so to find FF, you'd do mew*normalforce
the Greek letter is "mu", written μ off the sidebar (in advanced) or [itex]\mu[/itex]
"mew" is cuter though ;) so you should claim that it was deliberate even if it wasn't.

I got 1.857 m /s
That's a speed - you need an acceleration. Watch those units - it's not clear is it's just a typo or if you really did solve for balanced forces.


The usual starting point is [itex]\Sigma F_x = ma_x[/itex] and [itex]\Sigma F_y = ma_y[/itex] and notice that [itex]a_y=0 \Rightarrow \Sigma F_y = 0[/itex] ... doing it in order avoids confusion when you do really complicated systems.

The trick is to keep each direction separate when you write out these sums - OP already knows how to do friction, so there will be an N or FN or whatever in both equations. You get two equations and two unknowns - very cleanly.
 
  • #5
57
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yeah I meant to type 1.857m/ss, just forgot to hit it an extra time.
"ΣFx=max and ΣFy=may and notice that ay=0⇒ΣFy=0" , doesn't mean much to me. I'm cool with the coordinate system and all, but the way I was taught it was simpler then your formula
 
  • #6
Simon Bridge
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Hey no worries. The sigma just means "sum" - they tell you to sum all the forces in each direction. So you pick a direction to call positive, any force the other way is negative.

This way is simpler in the long run - where you have to sum a very large number of forces which are arbitrarily aligned. You exploited an intuitive understanding of the symmetry - which is good, don't get me wrong. Be aware that people who don't understand it so well may have trouble following you.

If you do it this way then you can just read off the forces in each direction - just write then down one after the other on one line and the correct sum will just "appear".

It's also good discipline to work out all the algebra before you substitute values. This is especially true for computer mediated work, because rounding errors accumulate and typos are easier when there are lots of decimals. Again, this is something that pays off more later on. The point of doing this stuff is so you can do harder problems later - it's "wax on wax off".

I was kinda hoping OP would notice the thing with the vector directions and realize <head slap> and/or get back to me for more info. It's difficult to balance how much to tell someone. Well see what happens.

note: ms-2 is often written m/s/s rather than m/ss even though that does not make much sense mathematically. It could also be m/(ss) or m/s2 ... some things that are easy to write on paper are annoying to do on a computer!
 

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