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m = 7.5 kg rests on the surface of a horizontal ta by a pulley system to another block which hangs

m = 7.5 kg rests on the surface of a horizontal ta by a pulley system to another block which hangs

Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter5: Newton's Laws Of Motion
Section: Chapter Questions
Problem 35PQ: A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If...
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4.5 please answer parts a-c

Problem 5: A block with mass mj = 7.5 kg rests on the surface of a horizontal table which has a coefficient of static friction of u, = 0.59. This block is connected by a pulley system to another block which hangs vertically. The hanging block has mass m2. Part (a) What is the minimum mass, m2.min, in kilograms, that will cause the system to move? Numeric : A numeric value is expected and not an expression. m2 min = Part (b) A second block with mass of 10 kg is placed on top of the first block. What is the new hanging mass, m2min', in kilograms, that will cause the system to move? Numeric : A numeric value is expected and not an expression. m2,min' = Part (c) The table is inclined at an angle, 0, relative to the horizontal in such a way that the pulley is at the highest point and the 10 kg is removed from m1. Write an equation for the magnitude of the normal force, FN, of the block. Expression : FN = Select from the variables below to write your expression. Note that all variables may not be required. T, cos(a), cos(4), cos(0), sin(a), sin(4), sin(0), a, µs, Q, 0, g, m1, m2, t Part (d) Write an equation for the tension, T, in terms of mass mj just before the block begins to slide. Note that the angle 0 is not so large that the block would slide down the table under its own weight (i.e. if the tension acting on mass m, were removed, the block would remain stationary based on the friction between the block and the table). Expression : T = Select from the variables below to write your expression. Note that all variables may not be required. cos(a), cos(4), cos(0), sin(a), sin(4), sin(0), a, µs, Q, 0, d, g, m1, m2, t Part (e) Write an equation for the minimum mass m, necessary for the block to slide when the table is inclined. Expression : 2,min = Select from the variables below to write your expression. Note that all variables may not be required. cos(a), cos(), cos(0), sin(a), sin(4), sin(0), a, µg, Q, 0, d, g, m1, m2, t Part (f) How is the minimum mass m, for the inclined table different from that for the horizontal table considered in part (a)? MultipleChoice: 1) It is greater for the inclined table S HOT Ciougn inrormation. 3) It is less for the inclined table. 4) It is the same as for the inclined table.
Transcribed Image Text:Problem 5: A block with mass mj = 7.5 kg rests on the surface of a horizontal table which has a coefficient of static friction of u, = 0.59. This block is connected by a pulley system to another block which hangs vertically. The hanging block has mass m2. Part (a) What is the minimum mass, m2.min, in kilograms, that will cause the system to move? Numeric : A numeric value is expected and not an expression. m2 min = Part (b) A second block with mass of 10 kg is placed on top of the first block. What is the new hanging mass, m2min', in kilograms, that will cause the system to move? Numeric : A numeric value is expected and not an expression. m2,min' = Part (c) The table is inclined at an angle, 0, relative to the horizontal in such a way that the pulley is at the highest point and the 10 kg is removed from m1. Write an equation for the magnitude of the normal force, FN, of the block. Expression : FN = Select from the variables below to write your expression. Note that all variables may not be required. T, cos(a), cos(4), cos(0), sin(a), sin(4), sin(0), a, µs, Q, 0, g, m1, m2, t Part (d) Write an equation for the tension, T, in terms of mass mj just before the block begins to slide. Note that the angle 0 is not so large that the block would slide down the table under its own weight (i.e. if the tension acting on mass m, were removed, the block would remain stationary based on the friction between the block and the table). Expression : T = Select from the variables below to write your expression. Note that all variables may not be required. cos(a), cos(4), cos(0), sin(a), sin(4), sin(0), a, µs, Q, 0, d, g, m1, m2, t Part (e) Write an equation for the minimum mass m, necessary for the block to slide when the table is inclined. Expression : 2,min = Select from the variables below to write your expression. Note that all variables may not be required. cos(a), cos(), cos(0), sin(a), sin(4), sin(0), a, µg, Q, 0, d, g, m1, m2, t Part (f) How is the minimum mass m, for the inclined table different from that for the horizontal table considered in part (a)? MultipleChoice: 1) It is greater for the inclined table S HOT Ciougn inrormation. 3) It is less for the inclined table. 4) It is the same as for the inclined table.
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