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Question
FnetA parallel to incline= F+mAgsin(60) - T
FnetA perpendicular to incline=mAgcos(60)
FnetB

Question

**Question**

FnetA parallel to incline= F+mAgsin(60) - T

FnetA perpendicular to incline=mAgcos(60)

FnetB parallel=T - mBgsin(30)

FnetB perpendicular= mBgcos(30)

Solution

5 (1 Ratings )

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**Question**

FnetA parallel to incline= F+mAgsin(60) - T

FnetA perpendicular to incline=mAgcos(60)

FnetB parallel=T - mBgsin(30)

FnetB perpendicular= mBgcos(30)

** **

**Solution**

Let's suppose the system is moving such that the mass mA moves down the incline

The component of the net force on the block A are

- mA a = T - mA g sin - F ---> T = F + mA g sin - mA a (1)

0 = NA - mA g cos ---> NA = mA g cos

The component of the net force on the block B are

- mB a = - T + mB g sin (2)

0 = NB - mB g ---> NB = mB g

Now, substitute equation (1) in equatio (2)

- mB a = - T + mB g sin

- mB a = - (F + mA g sin - mA a) + mB g sin

- mB a = - F - mA g sin + mA a + mB g sin

F + mA g sin - mB g sin = mA a + mB a

F + (mA - mB) g sin = mA a + mB a

find the acceleration of the system

a = [ F + (mA - mB) g sin ] / (mA - mB) (3)

Now, substitute equation (3) in equation (2)

- mB a = - T + mB g sin

T = mB a + mB g sin

T = mB ([ F + (mA - mB) g sin ] / (mA - mB)) + mB g sin

the tension is

T = [ mB F + mB (mA - mB) g sin ] / (mA - mB) + mB g sin

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