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Laws Of Motion

Question
CBSEENPH11016793

A body of mass m is placed on rough horizontal surface and pulled with force F at an angle straight theta as shown. Find the acceleration with which the body will accelerate.

Let coefficient of friction be μ.

Solution
Force is acting on the body as shown in the fig. below.
Now, breaking the rectangular component of F along horizontal and vertical direction.
(i) F cosθ in horizontal direction. (ii) F sinθ in vertically upward direction.
             
Weight of body mg acts in vertically downward direction.
Since the body is in equilibrium in vertical direction, 
∴              straight F space sinθ space plus space straight R space equals space mg 
rightwards double arrow                        straight R space equals space mg minus Fsinθ 
The force of limiting friction is given by, 
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#6 {main}</pre>
If straight F space cosθ space less than space straight f then acceleration will be zero, and
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#6 {main}</pre> then body will accelerate and is given by
                   space space straight a equals fraction numerator Fcosθ minus straight f over denominator straight m end fraction
                       equals fraction numerator Fcosθ space minus straight mu left parenthesis mg minus Fsinθ right parenthesis over denominator straight m end fraction
                       equals fraction numerator straight F left parenthesis cosθ plus μsinθ right parenthesis minus μmg over denominator straight m end fraction
                     space space space space equals fraction numerator straight F left parenthesis cosθ plus μsinθ right parenthesis over denominator straight m end fraction minus μg
So, 'a' is the acceleration with which the body will accelerate.