6. page 95, A5-kg mass is connected by a string to a 4-kg mass, and the string passes over a massless frictionless pulley so that the 5-kg mass sits on a tabletop and the 4-kg mass hangs straight down. If the system is static, what must be the frictional force acting between the tabletop and the 5-kg mass? AN at equilibrium 49 N Fe= 9.8x9-88.2 N >FF Umg 5-9.8=49 N 39.2 N √mg4•9.8=39.2 N

University Physics Volume 1
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Author:William Moebs, Samuel J. Ling, Jeff Sanny
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Chapter5: Newton's Law Of Motion
Section: Chapter Questions
Problem 86AP: A mass of 250.0 g is suspended from a spring hanging vertically. The spring stretches 6.00 cm. How...
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6.
page 95, A5-kg mass is connected by a string to a 4-kg mass, and the string passes over a massless
frictionless pulley so that the 5-kg mass sits on a tabletop and the 4-kg mass hangs straight
down. If the system is static, what must be the frictional force acting between the tabletop and
the 5-kg mass?
Gat equilibrium qN
49 N
la
F₂=9.8×9= 88.2 N
Umg
5-9.8-49 N
39.2 N
√mg 4•9.8=39.2 N
Transcribed Image Text:6. page 95, A5-kg mass is connected by a string to a 4-kg mass, and the string passes over a massless frictionless pulley so that the 5-kg mass sits on a tabletop and the 4-kg mass hangs straight down. If the system is static, what must be the frictional force acting between the tabletop and the 5-kg mass? Gat equilibrium qN 49 N la F₂=9.8×9= 88.2 N Umg 5-9.8-49 N 39.2 N √mg 4•9.8=39.2 N
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