how? Figure 5.3-1: Atwood\'s Machine assuming m2 > m1. a = (m2 - m1)g/m1 + m2 T

Question

how?

Figure 5.3-1: Atwood's Machine assuming m2 > m1. a = (m2 - m1)g/m1 + m2 T = 2m1m2g/m1 + m2 Figure 5.3-1 shows the free body diagrams and the resulting equations of motion that are found by applying Newton's second law of motion to the Atwood's machine. From the equations of motion, derive equations (5.2) and (5.3). (Hint: Think two equations and two unknowns.) Consider an Atwood's machine with m1 = (110.0 plusminus 0.1) g and m2 =(175.0 plusminus 0.1) g. Determine the acceleration of the masses.

Explanation / Answer

Apply N.L.M. ON BLOCK 2

For each body we have to choose a dirrection as positive to write the equation.

Generally this is direction in which object is accelerating

Here for block 2 this is downward direction

So forces in the downward direction are positive for block 2

m2g - T = m2a.........(equation1)

here we write summation of all forces with sign on one side and ma on the other which means the summation of all forces equals mass*acceleration

For block 1 it is accelerating up so upeards is positive direction

Forces in upeard direction are positive and downward negative

Force due to gravity is downwards so negative and tension is upwaards so positive

T - m1g = m1a........(equation2)

Add the two equations

m2g - m1g =( m1 + m2)*a

a = (m2 - m1)g/(m1 + m2)

Put the value of a in any eqquation and we get

T = m1( g + a).

put the value of a and we get

T = 2m1m2g/( m1+ m2)

For second part

m1 = 110 g = 0.11 kg

m2 = 175g = 0.175 kg

a = (0.175 - 0.11)*g/(0.11+0.175)

g= 9.8

So a = 2.235 m/s^2

error fraction in a = 2*(0.1/110 + 0.1/175)

= 0.00148

error in a = 0.00148*2.235=0.003

so acceleration = 2.235 +/- 0.003

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