1
Question
Block B has a mass m and is released from rest when it is on top of wedge A, which has a mass 3m. Determine the tension in cord CD while B is sliding down A? Neglect friction.
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Solution
The
free body diagram of block B is shown above
The
equations of motion are:
Perpendicular
to the incline: (1) N = m g cosq
Along
the incline: (2) mg sinq
- mN
= ma
Equations
of motion in the horizontal direction.
(3) T +
mNcosq
- Nsinq =
0
Substituting
(1) into (3) and solving for T, gives
(4) T = mgcosq
(sinq
- mcosq)
thus q
= 21.8 degress, m = 0.2 and
mB =5 kg and
the mass of B was increase
to mB = 25 kg while all other parameters remained the same.
Substituting these values into equation (3) give the following values
for T.
we get T = 8.45N.
The free body diagram of block B is shown above
The equations of motion are:
Perpendicular to the incline: (1) N = m g cosq
Along the incline: (2) mg sinq - mN = ma
Equations of motion in the horizontal direction.
(3) T + mNcosq - Nsinq = 0
Substituting (1) into (3) and solving for T, gives
(4) T = mgcosq (sinq - mcosq)
thus q
= 21.8 degress, m = 0.2 and
mB =5 kg and
the mass of B was increase
to mB = 25 kg while all other parameters remained the same.
Substituting these values into equation (3) give the following values
for T.
we get T = 8.45N.
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