- #1

- 13

- 0

## Homework Statement

Blocks A and B are connected by a cable that passes over support C. Friction between the blocks and the inclined surfaces may be neglected. Knowing that motion of block B up the incline is impending when m

_{B}= 4.5 kg, determine:

(a) the coefficient of static friction between the rope and the support

(b) the largest value of m

_{B}for which equilibrium is maintained

## Homework Equations

T

_{2}= T

_{1}e^(μ

_{s}β) (1)

β is in angle in radians

W = mg

## The Attempt at a Solution

My problem is that for part (b), my answer is wrong. I would like to know where I went wrong in my calculations. Thanks for the help in advance.

I found the forces working on Block A

ƩF

_{x}= 0

→ T

_{A}- W

_{A}sin(16) = 0

T

_{A}= 24.336 N

Then, Block B

ƩF

_{x}= 0

→ W

_{B}sin(16) - T

_{B}= 0

T

_{B}= 12.168 N

Then, I looked at support C where I found the angle to be 32°, and I used (1) to find the static friction.

β = (32)(∏/180) ≈ 0.559

T

_{A}= T

_{B}e^(μ

_{s}β)

24.336 N = (12.168 N)e^(μ

_{s}β)

e^(μ

_{s}β) = (24.336 N)/(12.168 N)

μ

_{s}β = ln [(24.336 N)/(12.168 N)]

μ

_{s}= (1/0.559)*ln [(24.336 N)/(12.168 N)]

μ

_{s}≈ 1.240

Then, since I know T

_{B}has to be greater than itself to upset the equilibrium, I did

T

_{B}= T

_{A}e^(μ

_{s}β)

T

_{B}= (24.336 N)e^[(1.240)(0.559)]

T

_{B}≈ 48.673 N

Then, to get the mass I divided the T

_{B}by g = 9.81

m

_{B}= (48.673)/(9.81) ≈ 4.962 kg

That is wrong.