Question #105196
5. A 15-kg object and a 10-kg object are suspended, joined by a cord that passes over a pulley with a radius of 15 cm and a mass of 3 kg. The cord has a negligible mass and does not slip on the pulley. Treat the pulley as a uniform disk, and determine the linear acceleration of the two objects after the objects are released from rest.

6. A block of mass m1 = 2 kg and a block of mass m2 = 6 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10 kg. These blocks are allowed to move on a fixed block-wedge of angle = 30 degrees. Draw free-body diagrams of both blocks and of the pulley. Determine (a) the acceleration of the two blocks, and (b) the tensions in the string on both sides of the pulley.
1
Expert's answer
2020-03-12T02:50:42-0400

As per the given question,

masses of the objects are (m1)=15kg(m_1)=15 kg

and (m2)=10kg(m_2) =10kg

mass of the pulley =3 kg

Radius of the pulley =15 cm= 0.15 m

now applying newton's law,

Let it is moving with a acceleration and 15kg block is hanged left side and 10 is hanged to right side of the pulley, Let the tension in the left side of the string is T1T_1 and right side of the string is T2T_2

So, 15gT1=15a(i)15g-T_1=15a-------(i)

T210g=10a(ii)T_2-10g=10a-------(ii)

from equation (i) and (ii)

15g10gT1T2=15a10a15g-10g-T_1-T_2=15a-10a

5g(T1T2)=5a5g-(T_1-T_2)=5a

T1T2=5g5a(iii)T_1-T_2=5g-5a--------(iii)

torque on the pulley,

(T1T2)R=Iα(iv)(T_1-T_2)R=I\alpha ----------(iv)

Now, from the equation (iii) and (iv)


(5g5a)R=mR22×α\Rightarrow (5g-5a)R=\dfrac{mR^2}{2}\times\alpha


5g5a=3a2\Rightarrow 5g-5a=\dfrac{3a}{2}


5g=5a+3a2\Rightarrow 5g= 5a+\dfrac{3a}{2}


13a2=5g\Rightarrow \dfrac{13a}{2}=5g


a=10g13m/sec2a=\dfrac{10g}{13} m/sec^2


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