Question

Can someone show me their work to find the answers to these problems? It's for my homework, but this screenshot shows the answers because it's a practice problem. I tried doing the practice problem with the answers to find out if i'm doing it the right way, but I can't get to the answers it's showing.

Answer 1

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!

Answer 2

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