Question

Problem 2: The system in the figure consists of two blocks (particles) A and B of masses ma -20 kg and тв--100 kg acted upon by a force P as shown. The coefficients of static and kinetic friction between blocks A and B is μ,- 4k = 0.5 There is no friction between block B and the ground. Gravity g9.81 m/s2 acts downward 20 kg 100 kg a) Specify appropriate position variables and write the equations of motion of the two particles considering the following 2 cases. For each case, clearly specify the additional equations/conditions on the position variables and/or the friction condition that is required to solve for the motion of the system. Also include any checks you would need to perform to ensure the solution obtained is appropriate. i) Block A slips towards the right relative to block B which is itself moving towards the right. ii) Block A does not move relative to block B and both blocks move towards the right. b) Determine the acceleration of each block for ii) P -40 N

Answer 1

(i) Draw the free body diagram when block A moves towards the right relative to Block B 2P N, For Block A 2P-MNm,a ....(1) For Block B. (Let block B also moves towards the right) F ma 1-тва в (ii) Draw the free body diagram when both blocks move towards the right. 2P

Σ Fx = ma 2P=(mA +1%)a ..。... (i) P- 60 N and block A moves towards the right relative to Block B From equations (1), (2), (3), and (4) 2x60-0.5x(m,g)-m,a, 120-0.5(20x9.81) 20x a, a.,-1.095 m/s2 0.5 (20×9.81)-100 a,-0.981 ln/s. P 60 N and both blocks move tovw s move towards the right From equation (5) 2P-(m,+m)a 2x 60 (20+100)a (ii) P-40 N and block A moves towards the right relative to Block B. 2x 40-0.5x(m,g)-m,a, 80一0.5(20×9.81)-20xaA a0.905 m/s P-40 N and both blocks move towards the right 2x40- (20+100)a 0.667 m/s2