two masses are connected by a rope of negligible mass, over a pulley with neglig
ID: 2186717 • Letter: T
Question
two masses are connected by a rope of negligible mass, over a pulley with negligible friction and negligible mass. The top mass is 4.0 kg and the hanging mass is 9.0kg The coefficient of sliding friction between the top block and the surface is 0.15. the angle of the surface is 30 degrees above the horizontal. Find the force of kinetic friction on the top block, and find the tension in the string and the magnitude of acceleration of the blocks. what i need is how to know how to do this, so please be as specific as possible. will rate if correctExplanation / Answer
Block 1 at the left moves downwards, dragged upwards by the string Tension T_1, while gravity acts downwards. Hence: m_1·a = m_1·g - T_1 Hence the tension in the cord between block1 and 2 is T_1 = m_1·(g - a) = 10kg · (9.81m/s² - 2.4m/s²) = 74.1N Block 2 in the middle moves to the left, dragged by T_1 to the left and T_2 to the right. Moreover there is a frictional force F = µ·m_2·g acting against the direction of motion. Therefore m_2·a = T_1 - T_2 - µ·m_2·g m_2·a = m_1·(g-a) - T_2 - µ·m_2·g µ = (m_1·g - (m_1+m_2)·a - T_2) / (m_2·g) Block 3 to the right moves upwhill, dragged by the string cord tension T_2, while a part of gravity and frictional force acting downhill. The normal force exerted by the incline to the block is N = m_3·g·cos(25°) The frictional force at block 3 is: F = µ·m_2·N = µ·m_3·g·cos(25°) Therefore m_3·a = T_2 - µ··m_3·g·cos(25°) -·m_3·g·sin(25°) µ = (T_2 - m_3·a - m_3·g·sin(25°)) / (m_3·g·cos(25°) equate the two equations for the friction coefficient and solve for T_2: (m_1·g - (m_1+m_2)·a - T_2) / (m_2·g) = (T_2 - m_3·a - m_3·g·sin(25°)) / (m_3·g·cos(25°) (m_1·g - (m_1+m_2)·a - T_2)·m_3·cos(25°) = (T_2 - m_3·a - m_3·g·sin(25°)) · m_2 T_2 = [(m_1·g - (m_1+m_2)·a)·m_3·cos(25°) + (a -g·sin(25°)) ·m_3·m_2] / [m_3·cos(25°) + m_2] = 22.5N (b) m_2·a = T_1 - T_2 - µ·m_2·g µ = (T_1 - T_2 - m_2·a ) / (m_2·g) = (74.1N - 22.5N - 4.1kg · 2.4m/s² ) / (4.1kg · 9.81m/s²) = 1.04Related Questions
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