Question

Dynamics, please answer all parts.Thank you

Given: Block A has a known constant mass of mA and block B has a known constant mass of ms. The coefficient of kinetic friction between block A and the incline is ex. Both block masses should be considered in your analysis, but the mass of block A is large enough so that block A slides down the incline after starting from rest. You may neglect the masses of the pulleys and cable. It can be shown that the kinematic (i.e., dependent motion) equations for the pulley system are: B 2sA + $B= C, where is a constant 2vA+ ve = 0, 2ax + as = 0, 2As A + Asp = 0, Let 0= arctan (3/4) Find: a) Using the equations above and your kinematics relations: If we assume the acceleration of block A has a constant value of 1 m/sdownward, what is the speed of the block B after it has moved up a distance of 1 ft. b) The free body diagram for both blocks. In other words, FULLY COMPLETE the diagrams shown below. c) Newton's 2nd Law equations of motion for each block in terms of mA, MB, fx, 0, g, etc. d) The acceleration of each block in terms of ma, ms, lk, 0, g, etc. Solution: Forces Inertial Terms Forces Inertial Terms FWA =

Answer 1

T Tx :) Hi 4 3 2T LA 4 ☺ AA = 2 m/s2 (9) VB 2 VA AB= 2aA ( agz 2 m/ve) L .305m ha I ft Vo ? 2x19.01+2) X.305 - Ve 2g+ajh [ Ve s. 2: 684. milsec c T (6) NB ap 27 for mag Sine t mag sine Mag. Cool my CSScanned with CamScanner

(C) mag sim Sino-at-for- iT-fo=MARA ( for Å block) [I- maga mean] : ( for 8' block) (2) magsime no que = aug Rund Na A mgcone) (d) add leq (1) 4 2x acq (2) 2T - 2 forume 2. mo ao ma mong sing -2T 46 mag Eoso = mna mag sing - amg memang Cone = 2 mo aq + MAMA the tutowe ** No=2an mag sine - amag - Me mag coso = (long & majan magsing -a meg ilk aming lone an = u mp + ma AB २ mag sing - 2 meg - lle ma glase ) umo TmA CScanned with CamScanner