Tutorial rcise Two forces F, and F, ast on a 5.30-kg object. F- 27.0 N and - 13.5 N. (a) Find the acceleration of the object for the configuration of forces shown in Figure (a). (b) Find the acceleration of the object for the configuration of forces shown in Figure (b). Part 1 of-Conceptualize Forces acting on an object add as vectors. The acceleration will be greater in magnitude in situation (b) than in situation (a). In boch cases, the acceleracion will be on the order of a few m/s, directed into the first quadrant. Part 2 of6-Categorie First, we will decermine the tocal force on the object. Since the object experiences acceleration, we can use the partide under net force model to desermine the acceleration with Newton's second law. Part 3 of 6- Analyze (a) Using the particle under a net force model, we calculace the net force on the object in the first figure - (2- 27 - 1, 135 ) I The following is the mathematical statement of Neton's second la Expressing the sum of the forces in vector notation, and substituting the value of the mass, we have EF - (2 27 - 138 13.5 )N= (5.30 5.3 ka) a Part4 of 6-Analyze Now, solving for the acceleration., gives 1x a- Your response dfers from the correct answer by more than 10%. Double check your calcularions. i+(1.01 Your response dffers from the correct answer by more than 10%. Double check your calculations. ) m/s". Remembering that a-Va- and substituting for the x and y components of acceleration, we have for the magnitude and drection m/r - m/s Oms) - (0 *- tan"
Tutorial rcise Two forces F, and F, ast on a 5.30-kg object. F- 27.0 N and - 13.5 N. (a) Find the acceleration of the object for the configuration of forces shown in Figure (a). (b) Find the acceleration of the object for the configuration of forces shown in Figure (b). Part 1 of-Conceptualize Forces acting on an object add as vectors. The acceleration will be greater in magnitude in situation (b) than in situation (a). In boch cases, the acceleracion will be on the order of a few m/s, directed into the first quadrant. Part 2 of6-Categorie First, we will decermine the tocal force on the object. Since the object experiences acceleration, we can use the partide under net force model to desermine the acceleration with Newton's second law. Part 3 of 6- Analyze (a) Using the particle under a net force model, we calculace the net force on the object in the first figure - (2- 27 - 1, 135 ) I The following is the mathematical statement of Neton's second la Expressing the sum of the forces in vector notation, and substituting the value of the mass, we have EF - (2 27 - 138 13.5 )N= (5.30 5.3 ka) a Part4 of 6-Analyze Now, solving for the acceleration., gives 1x a- Your response dfers from the correct answer by more than 10%. Double check your calcularions. i+(1.01 Your response dffers from the correct answer by more than 10%. Double check your calculations. ) m/s". Remembering that a-Va- and substituting for the x and y components of acceleration, we have for the magnitude and drection m/r - m/s Oms) - (0 *- tan"
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter5: Newton's Law Of Motion
Section: Chapter Questions
Problem 23P: While sliding a couch across a floor, Andrea and Jennifer exert forces FA and FJ on the couch....
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![Tutorial Exercise
Two forces F, and F, act on a 5.30-kg object. F, = 27.0 N and F, = 13.5 N.
90.0°
60.0
(a) Find the acceleration of the object for the configuration of forces shown in Figure (a).
(b) Find the acceleration of the object for the configuration of forces shown in Figure (b).
Part 1 of 6 - Conceptualize
Forces acting on an object add as vectors. The acceleration will be greater in magnitude in situation (b) than in situation (a). In both cases, the acceleration will be on the order of a few m/s?, directed into the first quadrant.
Part 2 of 6 - Categorize
First, we will determine the total force on the object. Since the object experiences acceleration, we can use the particle under net force model to determine the acceleration with Newton's second law.
Part 3 of 6 - Analyze
(a) Using the particle under a net force model, we calculate the net force on the object in the first figure.
EF = F, -F,
= (|27
27 i + 13.5
13.5 j| N
The following is the mathematical statement of Newton's second law.
EF = ma
Expressing the sum of the forces in vector notation, and substituting the value of the mass, we have
EF = (27 -
13.5 iN = (5.30
5.3 kg) a.
Part 4 of 6 - Analyze
Now, solving for the acceleration, gives
a = Your response differs from the correct answer by more than 10%. Double check your calculations. î + 1.91
Your response differs from the correct answer by more than 10%. Double check your calculations. j] m/s.
Remembering that a
= a = Va2 + a 2, and substituting for the x and y components of acceleration, we have for the magnitude and direction
]m/s) - ([
] m/s²)* = [
|m/s?
a =
e = tan
Submit
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Transcribed Image Text:Tutorial Exercise
Two forces F, and F, act on a 5.30-kg object. F, = 27.0 N and F, = 13.5 N.
90.0°
60.0
(a) Find the acceleration of the object for the configuration of forces shown in Figure (a).
(b) Find the acceleration of the object for the configuration of forces shown in Figure (b).
Part 1 of 6 - Conceptualize
Forces acting on an object add as vectors. The acceleration will be greater in magnitude in situation (b) than in situation (a). In both cases, the acceleration will be on the order of a few m/s?, directed into the first quadrant.
Part 2 of 6 - Categorize
First, we will determine the total force on the object. Since the object experiences acceleration, we can use the particle under net force model to determine the acceleration with Newton's second law.
Part 3 of 6 - Analyze
(a) Using the particle under a net force model, we calculate the net force on the object in the first figure.
EF = F, -F,
= (|27
27 i + 13.5
13.5 j| N
The following is the mathematical statement of Newton's second law.
EF = ma
Expressing the sum of the forces in vector notation, and substituting the value of the mass, we have
EF = (27 -
13.5 iN = (5.30
5.3 kg) a.
Part 4 of 6 - Analyze
Now, solving for the acceleration, gives
a = Your response differs from the correct answer by more than 10%. Double check your calculations. î + 1.91
Your response differs from the correct answer by more than 10%. Double check your calculations. j] m/s.
Remembering that a
= a = Va2 + a 2, and substituting for the x and y components of acceleration, we have for the magnitude and direction
]m/s) - ([
] m/s²)* = [
|m/s?
a =
e = tan
Submit
Skip (you cannot come back)
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