Example

Question:

What is the acceleration of the block and trolley system shown in Fig. 4.12(a), if the coefficient of kinetic friction between the trolley and the surface is \(0.04\)?
What is the tension in the string?
(Take \(g = 10\,\mathrm{m\,s^{-2}}\). Neglect the mass of the string.)

Solution:

As the string is inextensible and the pulley is smooth, the \(3\,\mathrm{kg}\) block and \(20\,\mathrm{kg}\) trolley both have the same acceleration.

Applying second law to the block (Fig. 4.12(b)): \[ 30 - T = 3a \] Applying second law to the trolley (Fig. 4.12(c)): \[ T - f_k = 20a \] Now, \(f_k = \mu_k N\), where \(\mu_k = 0.04\), \(N = 20 \times 10 = 200\,\mathrm{N}\): \[ f_k = 0.04 \times 200 = 8\,\mathrm{N} \] So, the equation for the motion of the trolley is: \[ T - 8 = 20a \quad \text{or} \quad T = 20a + 8 \] We already have \(30 - T = 3a\) so: \[ 30 - (20a + 8) = 3a \] \[ 22 = 23a \implies a = \frac{22}{23}\,\mathrm{m\,s^{-2}} \approx 0.96\,\mathrm{m\,s^{-2}} \] \[ T = 20a + 8 = 20 \times \frac{22}{23} + 8 = 27.13\,\mathrm{N} \] Thus, the acceleration of the system is \(0.96\,\mathrm{m\,s^{-2}}\) and the tension in the string is approximately \(27.1\,\mathrm{N}\).