Question

please help

2. Block A has a mass of 40 kg, and block B has a mass of8 kg. The coefficients of friction between all surfaces of contact are 0.2 and 4k 0.15. Determine (a) the accelerations of blocks A and B, (b) the tension in the cord, and (c) the velocities of blocks A and B after the block A has moved bv 0.5 m. Assume that initially the velocity of block A is 0 15° proach: . Isolate the body (bodies) of interest (free body) Draw your axis system (e.g., Cartesian, polar, path) Add in applied forces (e.g., weight, 225 lb pulling force) Replace supports with forces (e.g., normal force) Draw appropriate dimensions (usually angles for particles) . . Draw in the mass times acceleration of the particle; if unknown, do this in the positive . Write the equations of motion (i.e. Newton's Law for each body in each direction . Use kinematic equations to find positions, velocities, etc. direction according to your chosen axes, Solve for accelerations (and tension(s), vill supply the basic equations for this problem very soon Friction: 1 kV1 2 Newton's Law: Mass A in x-direction: 27-F1-Po cos(150)-N> sin( 15°) + T cos(15°) + mAg sin(25°) mAaA -2T -HN(T - HN2) cos (15°) - N2 sin(15°) + mag sin (250)-maaa(3) N,-N2 cos( 15°) + F2 sin( 150)-T sin(15°)-m,g cos(25°) = 0 N,-N, cos(15°) + (μ,W2-T) sin(15°) _ mAg cos(25°)-0 F2 cos (150) N2 sin(15°) - T cos(150) + mag sin(25°) maBr (HKN2 - T) cos(15°) +N2 sin(15°) +mBg sin (25)gagx N2 cos(15°)- F2 sin(15°) Tsin(15°) - mAg cos 25°) mgagy N> cos( 15°) + (T-AxN2) sin(15°)-m,g cos(250-mBaBy or Mass A in y-direction: or Mass B in X-direction: or ass B in y-direction: or Equations (3) to (6) have 6 unknowns, N1, N2, T, aA, aBx and agy. Constraint conditions provide two other relations Constraint conditions: inclined surface. The relative displacement, velocity and acceleration equations between A and If A will move by x in the x-direction, then B will move relative to A by 2x up the B are given as where Δ-4 and Δη are the absolute displacements of A and B, and Δη A is the relative displacement of B with respect to A. Other terms are defined in an analogues manner. Vectors Δ→ı and Δ→BA are defined as Æ = x→+0) and Δη.,--2xel, where i and j are the unit vectors along the x- and y-directions, and e, is the unit vector along the inclined surface (from left to right) between A and B. e, is given as e cos(-15°)i + sin(-15°)j- cos(150)7-s sin(15°)j. Thus, we have Δή, = xi-2x(cos(15°) i-sin(15°)) = x( 1-2 cos(15°))1+ 2x sin(15°) From which, it follows that VB = A (1-2 cos(15°))计2UA sin(15°) a,-CA(1-2 cos( 15°))ï + za, sin(15°) The last equation gives, ав,-a, (1-2 cos(15°)) aBy = 2a, sin(15°) (10) Equations (3)-(6), (9) and (10) are 6 equations in 6 unknowns. Solve them. Use an equation solver to solve them. This will save time. Also, if you make a mistake in the first round of calculations, then second and subsequent rounds of calculations will take much less time. You will find several equation solvers on the web Steps for subsequent calculations: 1. The above calculations give aa and T. 2. Equations (9) and (10) give aBx and aBy, respectively, and Eq. (8) gives äg. 4. Use Eq. (7) to compute vB

Answer 1

Tnitial velocity ak block A6 Multiply . li, in ee0 and、then adding in e2®, ae get on black A