Block on a Pulley: Determine Speed 2s Later

[KillerZ]

Homework Statement

At a given instant the 10-lb block A is moving downward with a speed of 6 ft/s. determine its speed 2s later. Block B has a weight of 4-lb, and the coefficient of kinetic friction between it and the horizontal plane is $$\o_{k} = 0.2$$. Neglect the mass of the pulleys and cord.

Homework Equations

$$v = v_{0} + a_{A}t$$

The Attempt at a Solution

FBD:

Block A:

$$+\downarrow \sum F_{y} = ma_{y}$$

$$10 - 2T = (0.311)(a_{A})$$

Block B:

$$+\rightarrow \sum F_{x} = ma_{x}$$

$$T - (0.2)(4) = (0.124)(a_{B})$$

$$+\downarrow \sum F_{y} = ma_{y}$$

$$4 - N = 0$$

Kinematics:

$$2S_{A} + S_{B} = l$$

$$2a_{A} + a_{B} = 0$$

$$2a_{A} = -a_{B}$$

Solving for T:

$$10 - 2T = (0.311)(a_{A})$$

$$10 - 2T = -(0.311)(a_{B})$$

$$a_{B} = -\frac{10 - 2T}{0.311} = \frac{T - (0.2)(4)}{0.124}$$

$$T = -16lb/2 = -8lb$$

$$a_{B} = -70.84ft/s^{2}$$

$$a_{A} = 83.72ft/s^{2}$$

$$v = v_{0} + a_{A}t$$

$$v = 6 + (83.72)(2) = 173.44 ft/s$$ I am not sure if this is right because that seems fast.

 

[willem2]

$$2a_{A} = -a_{B}$$

Solving for T:

$$10 - 2T = (0.311)(a_{A})$$

$$10 - 2T = -(0.311)(a_{B})$$

you forgot a factor (1/2) here.

$$a_{B} = -\frac{10 - 2T}{0.311} = \frac{T - (0.2)(4)}{0.124}$$

$$T = -16lb/2 = -8lb$$

T = -8 lb is not a solution of the above equaton. Try to give some more detail when you
try to solve for T again.
The negative tension in a rope should have tipped you off that something was wrong.

 

[KillerZ]

I didn't have a chance to get back to this but I finally solved it I think:

$$2a_{A} = -a_{B}$$

$$a_{A} = -\frac{a_{B}}{2}$$

$$10 - 2T = (0.311)(a_{A})$$

$$10 - 2T = -(0.311)(\frac{a_{B}}{2})$$

$$a_{B} = -\frac{20 - 4T}{0.311} = \frac{T - (0.2)(4)}{0.124}$$

$$-2.48 + 0.50T = 0.31T - 0.25$$

$$T = 2.769 lb$$

$$a_{A} = 14.37ft/s^{2}$$

$$a_{B} = 15.85ft/s^{2}$$

$$v = v_{0} + a_{A}t$$

$$v = 6 + (14.37)(2) = 34.73 ft/s$$

 

[willem2]

I didn't have a chance to get back to this but I finally solved it I think:

$$2a_{A} = -a_{B}$$

I'm sorry, this was an error I missed. a_A and a_B should have the same sign.

$$-2.48 + 0.50T = 0.31T - 0.25$$

$$T = 2.769 lb$$

this T is not the solution of the last equation. I get 0.19 T = 2.23, so T = 11.7
your also using only 2 digits of precision here, and then give the final answer with much
more digits.

 

[rl.bhat]

10 - 2T = 0.311(aA)
T - 0.8 = 0.124(aB)
aA = 2*aB
So 10 - 2T = 0.622*aB
T - 0.8 = 0.124*aB
So ( 10 - 2T)/(T - 0.8) = 0.622/0.124.
Now solve for T.