Answer to Question #143668 in Physics for Sebastian Luis

Question

A block of mass m1 = 2.00 kg and a block of mass m2 = 6.00 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 = 10.0 kg. The fixed, wedge- shaped ramp makes an angle of 휃 = 30° as shown in the figure. The coefficient of kinetic friction is 0.360 for both blocks. (a) Draw force diagrams of both blocks and of the pulley. Determine (b) the acceleration of the two blocks and (c) the tensions in the string on both sides of the pulley.

Answer 1

(a) The diagram:

(b) To find the acceleration, use Newton's second law (regular one with forces and the one for case of rotation). Assume positive x is to the right and y is to upward.

• Forces acting on m1:

$$T_1 − f_1 = m_1a,\\ N_1-m_1g=0.\\ T_1 = µm_1g + m_1a.$$

• Forces acting on m2:

$$−T_2 − f_2 + m_2g \text{ sin}θ = m_2a,\\ N_2-m_2g\text{ cos}\theta=0.\\ T_2 = m_2g \text{ sin}θ − µm_2g \text{ cos}θ − m_2a$$

• Torque on the pulley:

$$τ_\text{net} = Iα,\\ T_2R − T_1R = I\frac aR,\\\space\\ a =\frac{2R^2}{MR^2}(T_2 − T_1).$$

Express the acceleration:

$$a=\frac{2g[m_2 \text{ sin} θ − µ(m_2 \text{ cos} θ +m_1)]}{M + 2(m_1 + m_2)}=0.309\text{ m/s}^2.$$

(c) Using the equations for the objects of the system in part (b), find the tensions:

$$T_1 = m_1(µg + a)=7.674\text{ N},\\ T_2 = m_2(g \text{ sin} θ − µg\text{ cos} θ − a)=9.21\text{ N}.$$