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Question
A 2 kg object is subjected to three forces that give it an acceleration →a=−8 m/s2^i. If two of the three forces are F1=(30.0 N)^i+16.0 N)^j and F2=−(12 N)^i+(8 N)^j, find the third force.
A 2 kg object is subjected to three forces that give it an acceleration →a=−8 m/s2^i. If two of the three forces are F1=(30.0 N)^i+16.0 N)^j and F2=−(12 N)^i+(8 N)^j, find the third force.
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Solution
The correct option is B (−34 N)^i−(24 N)^j
Given: Acceleration of the body, →a=−8 m/s2 ^i
F1=(30 N)^i+(16 N)^j and
F2=−(12 N)^i+(8 N)^j
Net force acting on the object F=ma=2x(−8)^i=−(16 N)^i
Let the third force be F3=(x)^i+(y)^j
⇒F1+F2+F3=−(16 N)^i
⇒30^i+16^j−12^i+8^j+x^i+y^j=−16^i
⇒(30−12+x)^i+(16+8+y)^j=−16 ^i
⇒18+x=−16 and 24+y=0
⇒x=−34 N and y=−24 N
∴ The third force =F3=−(34 N)^i−(24 N)^j
Given: Acceleration of the body, →a=−8 m/s2 ^i
F1=(30 N)^i+(16 N)^j and
F2=−(12 N)^i+(8 N)^j
Net force acting on the object F=ma=2x(−8)^i=−(16 N)^i
Let the third force be F3=(x)^i+(y)^j
⇒F1+F2+F3=−(16 N)^i
⇒30^i+16^j−12^i+8^j+x^i+y^j=−16^i
⇒(30−12+x)^i+(16+8+y)^j=−16 ^i
⇒18+x=−16 and 24+y=0
⇒x=−34 N and y=−24 N
∴ The third force =F3=−(34 N)^i−(24 N)^j
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