A mass m1 = 5.7 kg rests on a frictionless table. It is connected by a massless

Question

A mass m1 = 5.7 kg rests on a frictionless table. It is connected by a massless and frictionless pulley to a second mass m2 = 3.9 kg that hangs freely.

3)

Now the table is tilted at an angle of = 69° with respect to the vertical. Find the magnitude of the new acceleration of block 1.

4) At what “critical” angle will the blocks NOT accelerate at all?

5) Now the angle is decreased past the “critical” angle so the system accelerates in the opposite direction. If = 21° find the magnitude of the acceleration.

6) Compare the tension in the string in each of the above cases on the incline:

A- T at 69° = Tcritical = T at 21°

B- T at 69° > Tcritical > T at 21°

C- T at 69° < Tcritical < T at 21°

Explanation / Answer

Since it is a massless and frictionless pulley, the tension in the string should be same throughout the length of the string.

Let it be T.

gravitational Force pulling m1 downwards is = m1gCos (since is the angle with the vertical, not horizontal)

So the acceleration of m1 = net force on m1 / mass of m1

a1 = (T - m1gCos )/ m1

acceleration of m2 , i.e a2 = net force on m2 / mass of m2 = ( m2g - T ) / m2

3.) Now, since the blocks are connected by a string their accelerations need to be the same

=> a1 = a2 or (T - m1gCos )/ m1 = ( m2g - T ) / m2

substituting the values,

(T - 5.7x9.8xCos 69 )/ 5.7 = ( 3.9x 9.8 - T ) / 3.9

(T - 20.018) x 3.9 = ( 38.22 - T) x 5.7

3.9T - 78.0702 = 217.854 - 5.7T

9.6 T = 295.9242

T = 30.825 N

now, a1 =  (T - m1gCos )/ m1 = (30.825 - 5.7 *9.8*Cos69 )/ 5.7 = 1.8955 m/s2

4.) At critical angle, block won't accelerate, so a1 = a2 = 0

a1 = (T - m1gCos )/ m1 =0

=> T = m1gCos

a2 = m2g - T = 0

=> T= m2g = 3.9*9.8 = 38.22 N

Therefore, T = m1gCos = m2g

=> m1gCos = m2g

m1Cos = m2

Cos = m2/m1= 3.9 / 5.7

= Cos -1 (3.9 /5.7) = 46.826 degrees

5.) The precodure remains the same. Hence by above derivations,

a1 = a2 or (T - m1gCos )/ m1 = ( m2g - T ) / m2

substituting the values,

(T - 5.7x9.8xCos 21 )/ 5.7 = ( 3.9x 9.8 - T ) / 3.9

(T - 52.1498) x 3.9 = ( 38.22 - T) x 5.7

3.9T - 203.3842 = 217.854 - 5.7T

9.6 T = 421.2382

T = 43.8789 N

now, a1 =  (T - m1gCos )/ m1 = (43.8789 - 5.7 *9.8*Cos21 )/ 5.7 = - 1.451 m/s2

The - ve sign just indicated the opposite direction of the acceleration.

6.) By looking at the results,

30.825 N < 38.22 N < 43.8789 N

so, C- T at 69° < Tcritical < T at 21°

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