Answer to Question #103645 in Classical Mechanics for Nichel Narayan

Question

Three objects are connected on a table as shown in the Figure.
The coefficient of kinetic friction between the block of mass m2
and the table is 0.350. The objects have masses of m1 = 4.00 kg,
m2 = 1.00 kg, and m3 = 2.00 kg, and the pulleys are frictionless.
(a) Draw a free-body diagram of each object. (b) Determine the
acceleration of each object, including its direction. (c) Determine
the tensions in the two cords. What If? (d) If the tabletop were
smooth, would the tensions increase, decrease, or remain the
same? Explain.

Answer 1

As per the question,

Coefficient of friction between the block $m_2$ and table $\mu$ =0.350

$m_1=4kg, m_2=1kg, m_3=2kg$

g=gravitational accelration

Let a be the net acceleration of the masses, N$(=m_2g=1g)$ bet the normal reaction on the table due to $m_2$ $T_1$ =tension between the string of block $m_1$ and $m_2$ and $T_2$ is the tension between the string of block $m_2$ and $m_3$

a) The FBD (free body diagram) is given below

b)

Force balance on the block $m_1$

Now, $m_1g-T_1=m_1a$

$4g-T_1=4a-------(i)$

force balance on the block $m_2$

$T_1-f_2-T_2=m_2a$

$T_1-\mu N-T_2=a$

$T_1-0.35g-T_2=a-----(ii)$

Force balance on the block $m_3$

$T_2-m_3g=m_3a$

$T_2-2g=2a-----(iii)$

adding equation (i) , (ii) and (iii)

$4g-0.35g-2g=7a$

$1.65g=7a$

$a=\dfrac{1.65g}{7}=0.24\times 9.8=2.4m/sec^2$

c)

Putting the value of a, in the equation (i) and (ii)

$T_1=4g-4a=4\times 9.8-4\times 2.4=29.6N$

$T_2=2a+2g=2(9.8+2.4)=24.4N$

d)

If friction force is negligible, then $f_s=0$

so, $a=\dfrac{4g-2g}{7}=\dfrac{2\times 9.8}{7}=2.8m/sec^2$

so, $T_1=4g-4a=34.4N$

$T_2=2a+2g=2\times 2.8+2\times 9.8=25.2N$

Hence tension,in both the string and the acceleration will increase.