Consider the system show with mA = 9.5 kg and mB =11.5kg.Theangles?A =59o and?B
ID: 2203801 • Letter: C
Question
Consider the system show with mA = 9.5 kg and mB =11.5kg.Theangles?A =59o and?B = 32o . a. In the absence of friction, what force F would be required to pull the masses at a constant velocity up the fixed inclines? b. The force F is now removed. What is the magnitude and direction of the acceleration of the two blocks? c. In the absence of F, what is the tension in the string?.Explanation / Answer
(a) Start by writing equations from Newton’s 2nd law for the forces acting on each mass separately. Since there is no friction, the forces perpendicular to the inclines are of no interest. For Ma (call it m1) the forces acting parallel to the 59.0° (?1) incline are the tension (T) upward and the component of its weight (-m1gsin?1) downward: ?F1 = 0 = T - m1gsin?1--------------->(1) For Mb (m1), the forces are the pulling force (F), the tension (-T, minus because it is downward), and the component of its weight down the incline (-m2gsin?2): ?F2 = 0 = F - T - m2gsin?2------------------->(2) Both (1) and (2) are equal to zero because the blocks are said to move at constant speed (that is, acceleration equals zero). Adding these together eliminates T and allows you to solve for F: 0 = F - m1gsin?1 - m2gsin?2 F = g(m1sin?1 + m2sin?2) = 9.80m/s²(9.50kgsin59.0° + 11.5kgsin32.0°) = 140N (rounded) (b) Here, we remove F (from equation one above) and set each equation equal to ma: ?F1 = m1a = T - m1gsin?1--------------->(3) And: ?F2 = m2a = - T - m2gsin?2------------------->(4) Adding together (3) and (4): m1a + m2a = - m1gsin?1 - m2gsin?2 a = -g(m1sin?1 + m2sin?2) / (m1 + m2) = -9.80m/s²(9.50kgsin59.0° + 11.5kgsin32.0°) / (9.50kg + 11.5kg) = -6.64m/s² (c) For which case (a) or (b) do want the tension? For (a) just use equation (1): 0 = T - m1gsin?1 T = m1gsin? = 9.50kg(9.80m/s²)sin59.0° = 79.8N For (b) use (3): m1a = T - m1gsin?1 T = m1a + m1gsin?1 = m1(a + gsin?1) = 9.50kg(-6.64m/s² + 9.80m/s²sin59.0°) = 16.7N