Created for CVCA Physics By Dick Heckathorn 6 December 2K+3 PROBLEMS p165 #2, #3 Created for CVCA Physics By Dick Heckathorn 6 December 2K+3

Problem Page 165 #2 A 40 kg block on a level, frictionless table is connected to a 15 kg mass by a rope passing over a frictionless pulley. What will be the acceleration of the mass?

Problem Page 165 #2 Ft on b Fe on b Fe on b 147 N Fnet a = = 147 N = 15 kg x 9.8 m/s2 2.7 m/s2 55 kg

What is the force on 40 kg mass? Where is the rest of the 147 N? What is the tension on the string? What is the force on 40 kg mass? 107 N Fon 40kg = 40 kg . 2.67 m/s2 107 N m . a F = 147 N a = 2.7 m/s2 Fon 15kg = 15 kg . 2.67 m/s2 m . a 40 N

Assumptions about Strings & Pulleys - 1 Strings are considered to have negligible mass and are capable of exerting only “pulling” forces on objects to which they are attached (tension forces).

Assumptions about Strings & Pulleys - 2 Strings transmit forces undiminished: the tension force in a string is the same throughout its length.

Assumptions about Strings & Pulleys - 3 A frictionless pulley changes the direction of a string without diminishing its tension.

Assumptions about Strings & Pulleys - 4 Strings are assumed not to stretch.

Problem Page 165 #3 A 3.0 kg mass is attached to a 5.0 kg mass by a string that passes over a frictionless pulley. When the masses are allowed to hang freely, what will be the acceleration of the masses and the magnitude of the tension in the string?

Problem Page 165 #3 Fe on b Fe on b Fnet 19.6 N Fnet a = 2.45 m/s2 = (5 – 3) kg x 9.8 m/s2 = 19.6 N 8 kg

Problem Page 165 #3 Fa Fs on b Fe on b Fstring = T = Fs on b + Fa T= 3.0 kg x 9.8 m/s2 +3.0 kg x 2.45 m/s2 T = 29.4 N + 7.35 N = 36.8 N

Problem Page 165 #3 36.8 N Fs on b3 Fa = 7.4 N F s on b 29.4 N Fe on b 29.4 N Fa = 12.2 N Fe on b 49.0 N Fs on b5 = FTotal on b5 = 61.2 N 36.8 N 61.2 N 49.0 N No No FTotal on b5 =

That’s All Folks!