Friction On 2 Blocks + Angled Pull

[sandman203]

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?

 

[TSny]

ok so I've changed ƩFyb = 0 = NB + 2×axsin15 - mb×g

The zero on the left of the equation is correct and the -m_bg on the right is correct.

However, the y-component of N_B is not N_B. 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 f_B which will be f_B sin15.

You can then use f_B = μ_sN_B to write the y-component of f_B as μ_sN_Bsin15. So, your equation \sumF_y = 0 will be an equation with just one unknown, N_B. Therefore you can find the value of N_B.

Then, you can set up \sumF_x = ma_x 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.

 

[sandman203]

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?

 

[TSny]

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 N_By is not the same as N_B. Using trig on the right triangle you get

N_By = N_Bcos(15)

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

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

f_By = f_Bsin(15) = μ_sN_Bsin(15)

You also know that the y component of the weight is

W_y = -mg


There is no y-component of acceleration, so Newton's 2nd law gives \sumF_y = 0. That is,

N_By + f_By + W_y = 0.

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

 

[sandman203]

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

 

[TSny]

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 f_B.

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 \sumF_x = f_Bx + N_Bx + W_x = m_Ba_x.

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 a_x.

 

[sandman203]

Ok so Nb = 17.481

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

so ax = -4.0422

Now subbing into fxab

F = 59.88N

 

[TSny]

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 a_x will be a positive number. You should not put a negative sign in front of a_x in the equation.

Note that N_B has an x-component that is negative. Also, are you sure you used the correct trig function to get the x-component of N_B? 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.

 

[sandman203]

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

 

[TSny]

Acceleration is correct.

F is not correct. Please show your work.

 

[sandman203]

Ok so N_A = 7×g - Fsin30
so F_fA = 0.6 × (7×g - Fsin30)

Now F_x = m×a_x = Fcos30 - F_fA

From before a_x = 2.803 & F_fA = 0.6 × (7×g - Fsin30)

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

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

F_x = 19.621 + 41.46 = Fcos30 + 0.6×Fsin30

F_x = 60.881 = F (cos30 + 0.6sin30)

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

F = 52.21N?

So i got this second time around.. I am not sure if i subtract F_fA or add it. Adding i get F = 38.053

 

[TSny]

That all looks very good! However, did you use the correct coefficient of friction?

 

[sandman203]

hmmmm 0.6? aren't we trying to overcome the static friction? I am not sure how i'd incorporate Kinetic friction because isn't the question asking the force needed to make the block slide from rest?

 

[TSny]

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.]

 

[sandman203]

Ah ok, yes the question does not state that block A is at rest. I am 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.

 

[sandman203]

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!

 

[TSny]

That's what I got. So, I believe it's correct. Good work!

 

[sandman203]

wow ok finally haha. guess we will see when it gets marked! thanks once again for the help dude realllllllllllly appreciate it!