Sure, give me a second and I’ll do the problems
So:
Ff = μsN
We know that the weight of block A is 46 N and block B is 27 N
So the net force on blocks A and C is equal to
Fac = ma ; since the acceleration is 0
Fac = 0
The net force on block b is
Fb(in the y direction) = ma
Since the acceleration is 0
T - mg = 0
T - 27 = 0
T = 27N
Now if we go back to blocks a and c we see that
ΣFx = 0 and ΣFy = 0
So the two forces acting in the x direction are tension and the frictional force between the blocks so...
ΣFx = T - Ff
and since ΣFx = 0 --> T - Ff = 0
Now Ff is equal to μsN so lets replace that with Ff
T - μsN = 0
Now to find the Normal force we have to look at the y direction:
ΣFy = N - mg (of a and c)
Since we know A and not C we can write
ΣFy = N - (46N + mgc)
ΣFy = 0
So --> N = 46N + mgc
Now we can plug this back into the other equation for net force in the x direction:
T - μs(46N + mgc) = 0 : Now we solved for tension earlier as it was 27 N and we have μs so:
27 - 0.20(46 + mgc) = 0
Subtract by 27
-0.20(46+mgc) = -27
Divide by -0.20
46 + mgc = 135
mgc = 89N ... Phew that was long! thats part a
Now part b)
C is lifted off now so there is no weight caused by C
Now net force on B in the y direction =
ΣFby = T - mg
Which equals ma, as now we have an acceleration
-ma = T - mg ; this acceleration is negative so a should be negative
We now mg is 27 as it is given so
-ma = T - 27N
Now we have to find the mass for the other side of the equation so
Divide the weight by 9.8 to get the mass as the weight is equal to mg
That equals 2.752 kg
Now we have:
-2.752a = T - 27N
Now add 27N to isolate for tension because we have to solve for it later
T = 27 - 2.752a
Lets leave that for now and move to block A to find the tension
ΣFx for block A = T - Ff (remember this is now kinetic friction!)
T - Ff = max (acceleration in the x direction)
Ff = μk N so...
T - μkN = max
Now notice that the normal force will be equal to the force on gravity (weight) on block A because there is no acceleration in the y direction so:
T - 46μk = max
Now we have to find the mass of block A and we use the same process as before for block b, divide the weight by g
46 / 9.81 = 4.689kg so
T - 46μk = 4.689 ax
We found the expression for tension earlier and we know μk so lets fill that in:
27 - 2.752a - 46(0.15) = 4.689a
Now lets group the a terms together so add -2.245a to the other side
27 - 46(0.15) = 7.441a
Simplify the left:
27 - 6.9
20.1 = 7.441a
Divide by 7.441
a = 2.701 : I hope it helped a bit!
a) tension in rope = weight of B = 27N
to prevent A from sliding, friction must be >= tensions in rope = 27N
friction = weight * coeff of frict.
= (weight of A+C) * 0.2
combining
27 = (46+ min wt. of C)*0.2
min wt. of C = 135 - 46
= 89N
b) if C is removed, friction force = 46*0.15 = 6.9N
let acceleration for A and B be x, rope tension be T n gravity be g=9.8
look at free body diag for A:
tension - friction = mass A * accel
T - 6.9 = (46/g)*x
T = (46/g)*x + 6.9
look at free body diag for B:
wt of B - tension = mass B * accel
27 - T = (27/g)*x
combining
27 - ( (46/g)*x + 6.9) = (27/g)*x
20.1 = ((27+46)/g)*x
x = 20.1*g/73
= 20.1*9.8/73
= 2.698
= 2.7 m/s^2
