Asked by Christina
Forces of 86 N east, 23 N north, 46 N south, 55 N west are simultaneously applied to a box of mass 13.8 kg. Find the magnitude of the box's acceleration?
Answers
Answered by
Elena
x: F(x) =86-55 = 31 N (east)
y: F(y) = 46-23 =23 N (south)
a(x) = F(x)/m =31/13.8 =2.25 m/s² (east)
a(y) = F(y)/m =23/13.8 =1.67 m/s² (south)
a= sqrt(a(x)² + a(y) ²) =2.8 1.67 m/s² (to the south of east)
y: F(y) = 46-23 =23 N (south)
a(x) = F(x)/m =31/13.8 =2.25 m/s² (east)
a(y) = F(y)/m =23/13.8 =1.67 m/s² (south)
a= sqrt(a(x)² + a(y) ²) =2.8 1.67 m/s² (to the south of east)
Answered by
Elena
The last line is
a= sqrt(a(x)² + a(y) ²) = sqrt(2.25²+1.67²)=
= 2.8 m/s² (to the south of east)
a= sqrt(a(x)² + a(y) ²) = sqrt(2.25²+1.67²)=
= 2.8 m/s² (to the south of east)
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