Answer
$Fx=Fcosθ =0.500×m×9.8m/s2×cos20∘=(4.6m)m/s2$
$fs,max=μsN=μs(mg−Fy)=μs(mg−Fsinθ)=0.600(m×9.8m/s2−0.500×m×9.8m/s2×sin20∘)=(4.9m)N$
The horizontal component of the force is less than the maximum value of static friction. Therefore, the block remains stationary. That is, acceleration a=0.
Work Step by Step
$F_{x}=F\cos\theta$$=0.500\times m\times9.8\,m/s^{2}\times\cos 20^{\circ}=(4.6m)\,m/s^{2}$
$f_{s,\,max}=\mu_{s}N=\mu_{s}(mg-F_{y})=\mu_{s}(mg-F\sin\theta)$$=0.600(m\times9.8\,m/s^{2}-0.500\times m\times9.8\,m/s^{2}\times\sin20^{\circ})$$=(4.9m)\,m/s^{2}$
Horizontal component of the force is less than the maximum value of static friction. Therefore, the block remains stationary. That is, acceleration a=0.
