A mass m 1 = 6.3 kg rests on a frictionless table. It is connected by a massless

Question

A mass m1 = 6.3 kg rests on a frictionless table. It is connected by a massless and frictionless pulley to a second mass m2= 3.2 kg that hangs freely.

1) What is the magnitude of the acceleration of block 1?

2) What is the tension in the string?

3) Now the table is tilted at an angle of = 75° 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 = 30° 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 75° = Tcritical = T at 30° b) T at 75° > Tcritical > T at 30° c)T at 75° < Tcritical < T at 30°

Explanation / Answer

acceleration of mass 1

a = F/m

a = m2*g/(m1 + m2)

a = 3.2*9.81/(6.3 + 3.2) = 3.304 m/sec^2

B. T = m1*a

T = 6.3*3.304

T = 20.82 N

C.

a = (m2*g - m1*g*sin A)/(m1 + m2)

a = (3.2*9.81 - 6.3*9.81*sin 15 deg)/(6.3 + 3.2)

a = 1.62 m/sec^2

D.

a = 0, when

m2*g = m1*g*sin A

sin A = m2/m1

Angle A = arcsin (3.2/6.3) = 30.53 deg

E.

a = (m1*g*sin A - m2*g)/(m1 + m2)

a = (6.3*9.81*sin 60 deg - 3.2*9.81)/(6.3 + 3.2)

a = 2.33 m/sec^2

F.

T at 75 deg

by C part

T = m2*g - m2*a

T = 6.3*9.81 - 6.3*1.62 = 51.597 N

T at critical angle

T = m2*g = 6.3*9.81 = 61.803N

T at 30 deg

T = m2*a + m2*g

T = 6.3*2.33 + 6.3*9.81 = 76.482 N

So option C is correct.

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