Answer to Question #143537 in Physics for Nielle

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 θ=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 is presented below:

(b) Since the string cannot stretch, the acceleration will be the same for both blocks. We will apply Newton's second law for block 1:

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

For block 2:

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

For our pulley:

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

Thus, we have three equations (the last ones for every object) and three undefined: tensions and acceleration. Solve this system, the solution will be

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

(c) For tension, the solution is

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