EXAMPLE 4.14 Connected Objects GOAL Use both the general method and the system a

Question

EXAMPLE 4.14 Connected Objects GOAL Use both the general method and the system approach to solve a connected two-body problem involving gravity and friction in PROBLEM (a) A block with mass m1 = 4.00 kg and a ball with mass m2 = 7.00 kg are connected by a light string that passes over a frictionless pulley, as shown in figure (a). The coefficient of kinetic friction between the block and the surface is 0.300. Find the acceleration of the two objects and the tension in the string. (b) Check the answer for the acceleration by using the system approach STRATEGY Connected objects are handled by applying Newton's second law separately to each object. The force diagrams for the block and the ball are shown in figure (b), with the +x-direction to the right and the +y-direction upwards. The magnitude of the acceleration for both objects has the same value, la 11 = 1a21 = a. The block with mass m1 moves in the positive x-direction, and the ball with mass m2 moves in the negative y- direction, so a 1 =-a2. Using Newton's second law, we can develop two equations involving the unknowns T and a that can be solved simultaneously. In part (b), treat the two masses as a single object, with the gravity force on the ball increasing the combined object's speed and the friction force on the block retarding it. The tension forces then become internal and don't appear in the second law (a) Two objects connected by a light string that passes over a frictionless pulley. (b) Force diagrams for the

Explanation / Answer

First Question -

(b) Write the expression for mass m1 –

T – uk*m1g = m1*a

T = m1*a + uk*m1g----------------------------------------------(i)

The expression for mass m2 is –

m2g – T = m2*a

=> T = m2*g - m2a---------------------------------------(ii)

Equalize (i) and (ii) –

m1*a + uk*m1g = m2*g -m2a

=> a*(m1+ m2) = m2g – uk*m1g

=> a = (m2 – uk*m1)*g / (m1 + m2)

Second question –

First Part:

By attaching additional mass to the ball, the downward acceleration of the ball shall increase.

Second Part:

New acceleration, a = 5.27 x 1.55 = 8.17 m/s^2

now put this value in the above expression –

8.17 = (m2 –0.30*4)*9.81 / (4 + m2)

=> 0.833*(4 + m2) = m2 – 1.2

=> 3.332 + m2*0.833 = m2 – 1.2

=> m2*0.167 = 4.532

=> m2 = 27.14 kg

So, the mass shall be increased to 27.14 kg.

Third Part:

For doubling the acceleration –

new acceleration, a = 5.27 x 2 = 10.54 m/s^2

put this value in the expression –

10.54 = (m2 –0.30*4)*9.81 / (4 + m2)

=> 1.07*(4 + m2) = m2 – 1.2

=> 4.28 + m2*1.07 = m2 – 1.2

=> m2*(1 – 1.07) =5.48

=> m2 = - 78.28 kg

This value of mass m2 is negative, which is not possible.

Therefore, doubling the existing acceleration is not possible.

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