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Friction On 2 Blocks + Angled Pull

  • Thread starter sandman203
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  • #26
TSny
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Im having trouble seeing how the acceleration to the left cause less normal force can someone please explain this to me, i really feel like im going insane over this. please help!
Please set up the equation [itex]\sum[/itex]Fy = mBay for the diagram I posted for B. That will allow you to find the numerical value of the normal force that A exerts on B.
 
  • #27
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ok so for block b ƩFy = 2×-gcos15 = -Fsin15 + NAB +2×axsin15??
 
  • #28
TSny
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Note [itex]\sum[/itex]Fy = may means

NBy + fBy + Wy = may where W is the force of gravity, fB is the friction force on B, and NB is the normal force.

Keep in mind that in the free body diagram for B, we are choosing the y-axis vertically. Does block B have any acceleration in the vertical direction?
 
  • #29
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ahh no it has none, otherwise it would lift off the block or go into block A. right?
 
  • #30
TSny
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Right. We are considering block B when it is just on the verge of slipping (but not yet slipping). So, yes, block B does not move vertically at all and ay = 0.

Now, how would you express NBy in terms of NB and the angle of 15 degrees?

How would you express fBy in terms of fB and the angle of 15 degrees?

How would you express Wy in terms of mB?
 

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  • #31
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ok so i've changed ƩFyb = 0 = NB + 2×axsin15 - mb×g

which decreases the normal.. i understand that now.. i didnt understand that before because i was thinking that only the x portion of the Force applied was acting on B. Now by realising the Y component of F also acts on B i can see how it reduces the normal.

So subbing into Fx of B i get an acceleration of 2.389s.

and subbing that into Fx equation for Block A i get 49.64N... is this correct?
 
  • #32
TSny
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ok so i've changed ƩFyb = 0 = NB + 2×axsin15 - mb×g
The zero on the left of the equation is correct and the -mbg on the right is correct.

However, the y-component of NB is not NB. You're going to need a trig function to get the y-component. (Study the free body diagram.)

I don't know how you got the term 2×axsin15. This term should represent the y-component of the friction force fB which will be fB sin15.

You can then use fB = μsNB to write the y-component of fB as μsNBsin15. So, your equation [itex]\sum[/itex]Fy = 0 will be an equation with just one unknown, NB. Therefore you can find the value of NB.

Then, you can set up [itex]\sum[/itex]Fx = max for block B and find the acceleration.

So subbing into Fx of B i get an acceleration of 2.389s.

and subbing that into Fx equation for Block A i get 49.64N... is this correct?
These answers are not correct, but they are not too far off.
 
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  • #33
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why is it not Nb, the normal is perpendicular to the incline surface? 2×axsin15 is the vertical component of F acting on B? or does only the horizontal portion of F work on B? how does friction come into action at all on the y-axis? i've changed the co-ord system when dealing with B. should i have not done that or something?
 
  • #34
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why is it not Nb, the normal is perpendicular to the incline surface?
Remember, we are choosing the y-axis vertical; that is, the y-axis is perpendicular to the ground and not perpendicular to the incline. Look at the attached figure and you can see that NBy is not the same as NB. Using trig on the right triangle you get

NBy = NBcos(15)

The figure also shows the y-component of the friction force, fBy. You can see that fBy = fBsin(15).

Since block B is on the verge of slipping, you can write fB = μsNB. So, that means

fBy = fBsin(15) = μsNBsin(15)

You also know that the y component of the weight is

Wy = -mg


There is no y-component of acceleration, so Newton's 2nd law gives [itex]\sum[/itex]Fy = 0. That is,

NBy + fBy + Wy = 0.

Substitute for each of these terms using the red-colored expressions above and solve for NB.
 

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  • #35
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Ok so using the above method i get Nb = 17.481, subbing into fxb i get ax = 2.529m/s^2 then subbing into Fxab i get 50.48N
 
  • #36
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Ok so using the above method i get Nb = 17.481
Yes, that looks correct. You can now easily calculate the value of the friction force fB.

subbing into fxb i get ax = 2.529m/s^2 then subbing into Fxab i get 50.48N
I don't believe those values are correct. You didn't show your work so I can't see where you went wrong.

Did you try setting up Newton's 2nd law for the x-component of the forces for block B?

You have the equation [itex]\sum[/itex]Fx = fBx + NBx + Wx = mBax.

Use the free body diagram for B to get expressions for the x-components of the forces and then use the equation to solve for ax.
 
  • #37
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Ok so Nb = 17.481

Fxb = 2*-ax = 17.481cos15 -2gsin15 + 10.4886cos15

so ax = -4.0422

Now subbing into fxab

F = 59.88N
 
  • #38
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Ok so Nb = 17.481

Fxb = 2*-ax = 17.481cos15 -2gsin15 + 10.4886cos15
We have chosen the positive direction of the x-axis to be to the left since we know that the blocks will accelerate to the left. Then ax will be a positive number. You should not put a negative sign in front of ax in the equation.

Note that NB has an x-component that is negative. Also, are you sure you used the correct trig function to get the x-component of NB? The x-component is the side opposite the 15 degree angle in the right triangle.

The weight force acts entirely along the negative y-axis. So, what is the x-component of the weight?

You have the correct expression for the x-component of the friction force.
 

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  • #39
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Ok so the x component of Wy = 0. So fxb = 2*ax = -17.481sin15 + 10.4886 cos15
so ax = 2.803m/s^2
& then F = 52.126N
 
  • #40
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Acceleration is correct.

F is not correct. Please show your work.
 
  • #41
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Ok so NA = 7×g - Fsin30
so FfA = 0.6 × (7×g - Fsin30)

Now Fx = m×ax = Fcos30 - FfA

From before ax = 2.803 & FfA = 0.6 × (7×g - Fsin30)

so Fx = 7×2.803 = Fcos30 - ((0.6 × 7 × g) - (0.6 × Fsin30)

Fx = 19.621 = Fcos30 - ((0.6 × 7 × g) - (0.6 × Fsin30)

Fx = 19.621 + 41.46 = Fcos30 + 0.6×Fsin30

Fx = 60.881 = F (cos30 + 0.6sin30)

Fx = F = 60.881 / (cos30 + 0.6sin30)

F = 52.21N?

So i got this second time around.. Im not sure if i subtract FfA or add it. Adding i get F = 38.053
 
  • #42
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That all looks very good! However, did you use the correct coefficient of friction?
 
  • #43
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hmmmm 0.6? aren't we trying to overcome the static friction? Im not sure how i'd incorporate Kinetic friction because isn't the question asking the force needed to make the block slide from rest?
 
  • #44
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We're trying to produce enough acceleration of A so that B will slip on A. Acceleration of A implies that A is slipping on the floor. So, you need to use the coefficient of kinetic friction for the friction between the floor and A.

[I didn't see where the problem stated the system starts from rest. If so, then you should check that the force necessary to overcome the static friction between A and the floor is not greater than the force required to accelerate the system at 2.80 m/s2 once A starts sliding. But you can check that the force that is just sufficient to start A slipping is not enough force to accelerate the system at 2.80 m/s2. So, applying just enough force to start A sliding would not cause B to slip on A.]
 
  • #45
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Ah ok, yes the question does not state that block A is at rest. Im going to go with the 0.4 co-efficient because u guys are way smarter than me. But can acceleration not imply that a body is being accelerated from rest? Anyways thanks a **** tonne for the help TSny, must of been a pain dealing with me lol.
 
  • #46
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so NA = 7×g - Fsin30
so FfA = 0.4 × (7×g - Fsin30)

Now Fx = m×ax = Fcos30 - FfA

From before ax = 2.803 & FfA = 0.6 × (7×g - Fsin30)

so Fx = 7×2.803 = Fcos30 - ((0.4 × 7 × g) - (0.4 × Fsin30)

Fx = 19.621 = Fcos30 - ((0.4 × 7 × g) - (0.4 × Fsin30)

Fx = 19.621 + 27.44 = Fcos30 + 0.4×Fsin30

Fx = 47.061 = F (cos30 + 0.4sin30)

Fx = F = 47.061 / (cos30 + 0.4sin30)

F = 44.14N please tell me this is finally correct!!
 
  • #47
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That's what I got. So, I believe it's correct. Good work!
 
  • #48
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wow ok finally haha. guess we will see when it gets marked!!! thanks once again for the help dude realllllllllllly appreciate it!!!!
 

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