Answered: F, and F, act on a 4.80-kg obje 90.0° O…

F, and F, act on a 4.80-kg obje 90.0° O Find the acceleration of the obje agnitude m/s? rection ° (counterd ) Find the acceleration of the obje agnitude m/s2 rection ° (counter

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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Question

F, and F, act on a 4.80-kg obje
90.0°
O Find the acceleration of the obje
agnitude
m/s?
rection
° (counterd
) Find the acceleration of the obje
agnitude
m/s2
rection
° (counter

Expert Solution

Step 1

Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and is in the direction of net force. Mass being the proportionality constant.

F=ma

Step 2

(a)

Magnitude of resultant force of ${\vec{F}}_{1} and {\vec{F}}_{2} is :$

$F = \sqrt{{|{\vec{F}}_{1}|}^{2} + {|{\vec{F}}_{2}|}^{2} + 2 |{\vec{F}}_{1}| |{\vec{F}}_{2}| \cos 90^{0}} F = \sqrt{{(30)}^{2} + {(10)}^{2}} F = 31.62 N$

Direction of resultant force :

$tanθ = \frac{|{\vec{F}}_{2}|}{|{\vec{F}}_{1}|} tanθ = \frac{10}{30} θ = \tan^{- 1} (\frac{10}{30}) θ = 18 . 43^{0}$

Use Newton's second law of force:

$F = ma$
$\Rightarrow a = \frac{F}{m} \Rightarrow a = \frac{31.62 N}{4.80 kg} \Rightarrow a = 6.58 m / s^{2}$

So acceleration of the object is $6.58 m / s^{2}$ in the direction of net force, i.e., at an angle of $18 . 43^{0}$ counterclockwise from ${\vec{F}}_{1}$

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