Answer:
The acceleration of the body is 7.0105 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
From the free body diagram of the given case
Vertical component of the applied force , [tex]F_{v}[/tex] = F cos(30°) = [tex]\dfrac{\sqrt{3}F }{2}[/tex] = 10[tex]\sqrt{3}[/tex]
Horizontal component of the applied force , [tex]F_{h}[/tex] = F sin(30°)
Balancing the normal force(N) from the wall and [tex]F_{h}[/tex] in horizontal direction.
N = [tex]F_{h}[/tex] = F sin(30°) = [tex]\dfrac{F}{2}[/tex] = 10 N
∵ The body is accelerating in vertical direction , the friction force acting on the body should be maximum.
∴ Maximum frictional force , f = k×N = 0.05×10 = 0.5 N
Let 'm' be the mass of the body and 'a' be the acceleration of the body in vertical direction and 'g' be the acceleration due to gravity.
∵ Body is accelerating in vertical direction , By Newton's 2nd law
[tex]F_{v}[/tex] - f - mg= ma
∴ 10[tex]\sqrt{3}[/tex] - 0.5 - (1×9.81) = 1×a
∴ a = 7.0105 [tex]\frac{m}{s^{2} }[/tex]
