Body A in the figure weighs 100 N, and body B weighs 59 N. The coefficients of f
ID: 2185619 • Letter: B
Question
Body A in the figure weighs 100 N, and body B weighs 59 N. The coefficients of friction between A and the incline are ?s = 0.54 and ?k = 0.21. Angle ? is 25Explanation / Answer
(a) When the system starts from rest, it will tend to move to the right (up the incline). Because the force of static friction is working to resist the motion of the block, it acts down the incline. For block A I'll use the subscript 1, and B will be 2. So you need to write equations for each block individually from Newton's second law. Friction, µ(s)w1cos?, is positiv, gravity (w1sin?) down the incline is positive, and the force of tension (T) in the cord is negative. The equation for block A ( w1) for the forces parallel to the incline is: m1a = w1sin? - T + µ(s)n-----> (1) n is the normal force which acts perpendicular to the incline. It can be found from the net forces on A, perpendicular to the incline: m2a = 0 = n - w1cos? n = w1cos?-----> (2) Here, w1cos? is the component of gravity perpendicular to the incline. Substituting (2) into (1): m1a = w1sin? - T + µ(s)w1cos?-----> (3) Now, the net forces on B (which are only vertical) are tension(T, and has the same value as T in the equations above, but opposite in direction, therefore it is positive) and weight (w2). The equation for B is: m2a = T - w2 ------> (4) Adding (3) and (4) together eliminates T, you can then solve for acceleration: m1a + m2a = [w1sin? - T + µ(s)w1cos?] + (T - w2 ) (m1+ m2)a = w1sin? - w2 + µ(s)w1cos? a = [ w1sin? - w2 + µ(s)w1cos?] / (m1+ m2)-----> (5) The mass of A (m1) can be found from: w1= m1g m1= w1/ g = 100N / 9.8m/s² = 10kg Block B then is 7.0kg by the same method. Now plug these masses along with the other data you have into (5) to find acceleration for part (a): a = [100Nsin26° - 69N + 0.58(100N)cos26°] / (10kg + 7.0kg) = 1.6m/s² So block A accelerates down the incline when it starts from rest. (b) In this case, friction acts downward, so it is positive, and since it is moving, it is kinetic friction. The other forces continue to act in the same direction for block A. Since A is moving up the incline, B is dropping (but take down as positive as it is in the diretion of motion). The equation for B is as (2) in part (a) The equation for A is: m1a = w1sin? - T + µ(k)w1cos? Thus, equation (5) becomes: a = [w1sin? - w2+ µ(k)w1cos?] / (m1+ m2) = [100Nsin26° - 69N + 0.28(100N)cos26°] / 17kg = 2.0 x 10?4m/s² So A moves up the incline while its acceleration acts opposite of it’s motion tending to slow it down and bring it to rest. (c) The trend then is that friction acts in the opposite direction of motion, so for motion of block A down the incline: a = [w1sin? - w2 - µ(k)w1cos?] / (m1+ m2) = [100Nsin26° - 69N - 0.28(100N)cos26°] / 17kg = -3.0m/s² So block A moves down the incline slowing down because the acceleration acts up the incline.