Question

Part A moves 2 ft up the inclino. The Determine the velocity of the 47 lb block A if the two blocks are released from rest and the 22 lb block coefficient of kinetic friction between both blocks and the inclined planes is 14 = 0.10 (Figure 1) Figure < 1 of 1 > Express your answer with the appropriate units. HAO? Value Units Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Provide Feedback Next >

Answer 1

The free-body diagram for block A is

т - - + Д mag sino a — та озо“ 40°

From the diagram

N = mag cos 60°

We know

f% = μ.N

fk= him ag cos 60°

The equation of motion for block A is

mag sin 60° – 27 – fk=m40A

m4g sin 60° – 2T - Mm 49 cos 60° = m 40A

Now the free body diagram for the block B is

Atomoo mog singo t . msg(0530° r 30°

From the diagram

N = mBg cos 30°

Therefore

f = μέN' = μm Bg coς 30°

The equation of motion of the block B is

T - mbg sin 30° – f= mbab

→T – mog sin 30º – Mambg cos 30º = mbag

Multiplying both sides by 2 we get

2T – 2m og sin 30° – 2umbg cos 30º = 2mgab

Adding equation 1 to it we get

2T – 2mBg sin 30° -2umbg cos 30° +mag sin 60° – 2T - HRM Ag cos 60º = 2mgas + maa

> mag sin 60°-HRM Ag cos 60°—2m og sin 30°—2 km Bg cos 30º = 2m gas+ m44

Now, as the string is inextensible, the two acceleration are related by

ag = 24

and the distance traveled by the two blocks can be related by

dg = 2dA

- It

Putting it in the above equation

> mag sin 60°-HRM Ag cos 60°—2m og sin 30°—2 km Bg cos 30º = 4mga A+ m44

→ m4g sin 60°– Hm Ag cos 60°—2mbgsin 30°—24MBg cos 30º = (4mb+ та)ая

>> a = (mA (sin 60° – Hk cos 60°) – 2mb(sin 30º – My cos 30°)), (4mb+ma)

Now, putting the values in the equation in the proper units we get

a = (47(sin 60° – 0.1 cos 60°) – 2 x 22 sin 30º – 0.1 cos 30)32.174 (4 x 22 +47)

>> DA= (38.353 – 18.189)32.174 != 4.8055 ft/s 135

The distance traveled by block A is

- It

We know

= +2a Ada

= = 02 +2 x 4.8055 x 1

va= V2 X 4.8055 x1= 3.1 ft/s

So, the velocity of the block A is 3.1 ft/s.