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Normal force on a block on an incline

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Homework Statement


In the figure, a crate of mass m = 107 kg is pushed at a constant speed up a frictionless ramp (θ = 32°) by a horizontal force [URL]http://edugen.wiley.com/edugen/shared/assignment/test/session.quest1510980entrance1_N10037.mml?size=14&rnd=1298229739658 [Broken][/URL].[/URL] The positive direction of an x axis is up the ramp, and the positive direction of a y axis is perpendicular to the ramp. (a) What is the magnitude of [URL]http://edugen.wiley.com/edugen/shared/assignment/test/session.quest1510980entrance1_N10037.mml?size=14&rnd=1298229739658 [Broken][/URL]?[/URL] (b) What is the magnitude of the normal force on the crate?

http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c05/q32.jpg


2. The attempt at a solution

I just wanted to see if this is right, I am very confused on Newton's Laws but from staring at this for hours I came up with:

F(perpendicular)= m.g.cosθ= Fy
Fy= (107kg)(9.8 m/s^2)cos(32)=72.2 N which would be the answer to part (b)? Because the crate is not moving in the y direction, the forces must cancel each other out and the normal force acting on the crate would be 72.2 N?

for part (a):
F(parallel)= m.g.sinθ= Fx
Fx= (107kg)(9.8m/s^2)sin(32)= 36.8 N
Since the crate is also not moving in the x direction (up the slope), the force pushing the on the crate, [URL]http://edugen.wiley.com/edugen/shared/assignment/test/session.quest1510980entrance1_N10037.mml?size=14&rnd=1298229739658[/URL] would have to equal the force that is acting on the crate in the -x direction, in other words the answer to part (b) would be 36.8 N?

I'm out of attempts on my homework and wanted to make sure it was okay before I entered the solution, any feedback would be wonderful! Thanks c:
 
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Answers and Replies

  • #2
Doc Al
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F(perpendicular)= m.g.cosθ= Fy
Fy= (107kg)(9.8 m/s^2)cos(32)=72.2 N which would be the answer to part (b)? Because the crate is not moving in the y direction, the forces must cancel each other out and the normal force acting on the crate would be 72.2 N?
While it's true that ΣFy = 0, the normal force is not simply mgcosθ. (The applied force F affects the normal force, since it has a y-component.)

for part (a):
F(parallel)= m.g.sinθ= Fx
Fx= (107kg)(9.8m/s^2)sin(32)= 36.8 N
Since the crate is also not moving in the x direction (up the slope), the force pushing the on the crate, [URL]http://edugen.wiley.com/edugen/shared/assignment/test/session.quest1510980entrance1_N10037.mml?size=14&rnd=1298229739658[/URL] would have to equal the force that is acting on the crate in the -x direction, in other words the answer to part (b) would be 36.8 N?
I assume part a asks for the applied force F? If so, realize that ΣFx = 0, since the block moves with constant speed. What's the x-component of F? (Realize that F is horizontal, not parallel to the slope.)
 
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  • #3
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While it's true that ΣFy = 0, the normal force is not simply mgcosθ. (The applied force F affects the normal force, since it has a y-component.)
Do you mean that the applied force would be sqrt[(mgcosθ)^2 - (mgsinθ)^2] then? Sorry I'm still confused.

I assume part a asks for the applied force F? If so, realize that ΣFx = 0, since the block moves with constant speed. What's the x-component of F? (Realize that F is horizontal, not parallel to the slope.)
Yes, part a asks for the applied force; The x component of F is mgsinθ = 36.8 N?
 
  • #4
Doc Al
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Do you mean that the applied force would be sqrt[(mgcosθ)^2 - (mgsinθ)^2] then? Sorry I'm still confused.
No. In any case, you're solving for the normal force here, right? Solve part a first.

Yes, part a asks for the applied force; The x component of F is mgsinθ = 36.8 N?
No, mgsinθ is the x component of the weight.

Do this. Identify all forces acting on the crate. There are three.

Express the x and y components of those forces in terms of θ. Write an equation expressing ΣFx = 0. Then you can solve for F.

Next write an equation for ΣFy = 0. Then you can solve for the normal force.
 
  • #5
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No. In any case, you're solving for the normal force here, right? Solve part a first.

No, mgsinθ is the x component of the weight.

Do this. Identify all forces acting on the crate. There are three.

Express the x and y components of those forces in terms of θ. Write an equation expressing ΣFx = 0. Then you can solve for F.

Next write an equation for ΣFy = 0. Then you can solve for the normal force.
The x component of the weight? I thought w=mg only had a y component... the forces acting on the crate are the normal force, the force of gravity, and the applied force.

for F(normal):
Fx= F(parallel)cosθ= F(parallel)cos(32)
Fy= F(parallel)sinθ= F(parallel)sin(32)

for F:
Fx= (parallel)cos(-θ)= (parallel)cos(-32)
Fy= (-Fny) sin(-θ)= (-Fny)sin(-32) (Fny is the y component of the normal force)

for F(gravity)
Fx= 0
Fy= mg(sin(θ))= (81kg)(sin32)

for F then:
∑Fx= Fx(normal)+Fx(applied)+Fx(gravity)=0
∑Fy=Fy(normal)+Fy(applied)+Fy(gravity)=0



:confused:
 
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  • #6
Doc Al
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The x component of the weight? I thought w=mg only had a y component...
For some reason I thought you were taking x as parallel to the incline. My bad! So you are taking x and y to be the horizontal and vertical. That's fine.
the forces acting on the crate are the normal force, the force of gravity, and the applied force.
Good.

It's confusing when you use the same symbol (F) for all the forces. I'd use:
F for the applied force.
W = mg for the weight.
N for the normal force.

for F(normal):
Fx= F(parallel)cosθ= F(parallel)cos(27)
Fy= F(parallel)sinθ= F(parallel)sin(27)
I assume this is supposed to be the normal force? (I thought the angle was 32 degrees?)
If you call the normal force N, then:
Nx = -Nsinθ
Ny = Ncosθ

for F:
Fx= (parallel)cos(-θ)= (parallel)cos(-27)
Fy= (-Fny) sin(-θ)= (-Fny)sin(-27) (Fny is the y component of the normal force)
Is this the applied force F? It's horizontal, so it has no y component.

for F(gravity)
Fx= 0
Fy= mg(sin(θ))= (81kg)(sin32)
Since gravity acts vertically, Fgy = mg, not mgsinθ.

for F then:
∑Fx= Fx(normal)+Fx(applied)+Fx(gravity)=0
∑Fy=Fy(normal)+Fy(applied)+Fy(gravity)=0
OK, but you need to plug in the values for the forces.

Give it another try.
 
  • #7
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For some reason I thought you were taking x as parallel to the incline. My bad! So you are taking x and y to be the horizontal and vertical. That's fine.

Good.

It's confusing when you use the same symbol (F) for all the forces. I'd use:
F for the applied force.
W = mg for the weight.
N for the normal force.


I assume this is supposed to be the normal force? (I thought the angle was 32 degrees?)
If you call the normal force N, then:
Nx = -Nsinθ
Ny = Ncosθ


Is this the applied force F? It's horizontal, so it has no y component.


Since gravity acts vertically, Fgy = mg, not mgsinθ.


OK, but you need to plug in the values for the forces.

Give it another try.
Thank you very much for putting so much time into helping me!

Yeah the angle θ should be 32, sorry I got it mixed up with another problem... originally my x axis was along the incline (aka parallel) but I'm having too much trouble visualizing it so I am going to set it up so that W is -y direction, and F is the +x direction.


Okay so like you said, (with #s):

Nx=-Nsinθ = -Nsin(32)= -N(0.53) (we don't know N, it's what we're solving for)
Ny=Ncosθ = Ncos(32)= N(0.85)


Fx= F(applied) (which we want to solve for so we don't know)
Fy=0


Wx=0
Wy=mg= (81kg)(9.8m/s^2)=793.8N

To solve, we would set up the equations:
∑Fx= Nx+Fx+Wx=0
∑Fy=Ny+Fy+Wy=0

∑Fx= (0.53)N+Fx+0=0
∑Fy= (0.85)N+0+793.8=0

Er...
0.85N=-793.8
N=-933.9 Newtons

and plugging in,
(-933.9)(.53)+Fx=0
Fx=495 Newtons?
 
  • #8
Doc Al
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Yeah the angle θ should be 32, sorry I got it mixed up with another problem... originally my x axis was along the incline (aka parallel) but I'm having too much trouble visualizing it so I am going to set it up so that W is -y direction, and F is the +x direction.
OK.

Okay so like you said, (with #s):

Nx=-Nsinθ = -Nsin(32)= -N(0.53) (we don't know N, it's what we're solving for)
Ny=Ncosθ = Ncos(32)= N(0.85)
I wouldn't plug in actual numbers until the last step. That will eliminate some rounding off errors.


Fx= F(applied) (which we want to solve for so we don't know)
Fy=0


Wx=0
Wy=mg= (81kg)(9.8m/s^2)=793.8N
OK, except that the weight acts down, so Wy should be negative.

To solve, we would set up the equations:
∑Fx= Nx+Fx+Wx=0
∑Fy=Ny+Fy+Wy=0
OK.

∑Fx= (0.53)N+Fx+0=0
∑Fy= (0.85)N+0+793.8=0
You made some sign errors. (That's why you get a negative answer for N.)

Er...
0.85N=-793.8
N=-933.9 Newtons
OK, but positive. And if you evaluate N = mg/cos(32), you'll get a more accurate answer.

and plugging in,
(-933.9)(.53)+Fx=0
Fx=495 Newtons?
Good. (But same comment about rounding off errors.)
 
  • #9
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OK.
I wouldn't plug in actual numbers until the last step. That will eliminate some rounding off errors.
OK, except that the weight acts down, so Wy should be negative.
OK.
You made some sign errors. (That's why you get a negative answer for N.)
OK, but positive. And if you evaluate N = mg/cos(32), you'll get a more accurate answer.
Good. (But same comment about rounding off errors.)
Oops, so N should be (as you said) mg/cosθ= (107kg)(9.8m/s^2)/cos(32)=1236.5 Newtons

for F then that would mean:
Nsin(32)+Fx=0
[(107)(9.8)(sin(32))/cos(32)]+Fx=0
Fx= -(107)(9.8)tan(32)= -655.2 Newtons
 
  • #10
Doc Al
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Oops, so N should be (as you said) mg/cosθ= (107kg)(9.8m/s^2)/cos(32)=1236.5 Newtons
Good. (You had the wrong mass earlier, for some reason.)

for F then that would mean:
Nsin(32)+Fx=0
[(107)(9.8)(sin(32))/cos(32)]+Fx=0
Fx= -(107)(9.8)tan(32)= -655.2 Newtons
Again, a sign error, but otherwise perfect.

-Nsin(32) + Fx = 0 → Fx = Nsin(32) = mgtan(32)
 
  • #11
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Good. (You had the wrong mass earlier, for some reason.)


Again, a sign error, but otherwise perfect.

-Nsin(32) + Fx = 0 → Fx = Nsin(32) = mgtan(32)
Thank you so very very very much. You helped me out a ton! I'm really glad I got to work through this problem with you!
 

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