As your plane circles an airport, it moves in a horizontalcircle of radius 2100

Question

As your plane circles an airport, it moves in a horizontalcircle of radius 2100 m with a speed of 380 km/h. If the lift ofthe airplane's wings is perpendicular to the wings, at what angleshould the plane be banked so that it doesn't tend to slipsideways? In the solution below I was wondering, where does the(5/18)m/s come from and how you would go about finding this number?I understand the rest of the solution I was just unsure aboutthis one part. Giventhat Radius of thehorizontal circle is (r) = 2100m Speed of theplane (v) = 380km/h                                   = 380*(5/18)m/s                                   = 105.5m/s Acceleration dueto gravity(g) =9.81m/s2      If we choose the positive y axis topoint vertically upward and the x direction to point toward thecenter of the circle, then N is at an angle since it isperpendicular to the banked surface. This angle is the exact theta wewould be looking for to represent the banking angle of theroadway. Thus, N has horizontal and verticalcomponents       (N = (Nsin)x + (Ncos)y)and W = -Wy = -mgy First, we can determine N from thecondition Fy = 0:               Fy = Ncos-W = 0                  N = W/cos = mg/cos.................(1) And Fx =Nsin =max =macp =mv2/r Thus, Nsin =(mg/cos)sin =mv2/r        ( from equation (1) ) The plane banked with an angleis         tan = v2/gr Or =tan-1(v2/gr)       =tan-1 ([105.55m/s]2/[9.81m/s2][2100m])         =tan-1 (0.5408)        =28.41         =280                                As your plane circles an airport, it moves in a horizontalcircle of radius 2100 m with a speed of 380 km/h. If the lift ofthe airplane's wings is perpendicular to the wings, at what angleshould the plane be banked so that it doesn't tend to slipsideways? In the solution below I was wondering, where does the(5/18)m/s come from and how you would go about finding this number?I understand the rest of the solution I was just unsure aboutthis one part. Giventhat Radius of thehorizontal circle is (r) = 2100m Speed of theplane (v) = 380km/h                                   = 380*(5/18)m/s                                   = 105.5m/s Acceleration dueto gravity(g) =9.81m/s2      If we choose the positive y axis topoint vertically upward and the x direction to point toward thecenter of the circle, then N is at an angle since it isperpendicular to the banked surface. This angle is the exact theta wewould be looking for to represent the banking angle of theroadway. Thus, N has horizontal and verticalcomponents       (N = (Nsin)x + (Ncos)y)and W = -Wy = -mgy First, we can determine N from thecondition Fy = 0:               Fy = Ncos-W = 0                  N = W/cos = mg/cos.................(1) And Fx =Nsin =max =macp =mv2/r Thus, Nsin =(mg/cos)sin =mv2/r        ( from equation (1) ) The plane banked with an angleis         tan = v2/gr Or =tan-1(v2/gr)       =tan-1 ([105.55m/s]2/[9.81m/s2][2100m])         =tan-1 (0.5408)        =28.41         =280                                Giventhat Radius of thehorizontal circle is (r) = 2100m Speed of theplane (v) = 380km/h                                   = 380*(5/18)m/s                                   = 105.5m/s Acceleration dueto gravity(g) =9.81m/s2      If we choose the positive y axis topoint vertically upward and the x direction to point toward thecenter of the circle, then N is at an angle since it isperpendicular to the banked surface. This angle is the exact theta wewould be looking for to represent the banking angle of theroadway. Thus, N has horizontal and verticalcomponents       (N = (Nsin)x + (Ncos)y)and W = -Wy = -mgy First, we can determine N from thecondition Fy = 0:               Fy = Ncos-W = 0                  N = W/cos = mg/cos.................(1) And Fx =Nsin =max =macp =mv2/r Thus, Nsin =(mg/cos)sin =mv2/r        ( from equation (1) ) The plane banked with an angleis         tan = v2/gr Or =tan-1(v2/gr)       =tan-1 ([105.55m/s]2/[9.81m/s2][2100m])         =tan-1 (0.5408)        =28.41         =280 If we choose the positive y axis topoint vertically upward and the x direction to point toward thecenter of the circle, then N is at an angle since it isperpendicular to the banked surface. This angle is the exact theta wewould be looking for to represent the banking angle of theroadway. Thus, N has horizontal and verticalcomponents       (N = (Nsin)x + (Ncos)y)and W = -Wy = -mgy First, we can determine N from thecondition Fy = 0:               Fy = Ncos-W = 0                  N = W/cos = mg/cos.................(1) And Fx =Nsin =max =macp =mv2/r Thus, Nsin =(mg/cos)sin =mv2/r        ( from equation (1) ) The plane banked with an angleis         tan = v2/gr Or =tan-1(v2/gr)       =tan-1 ([105.55m/s]2/[9.81m/s2][2100m])         =tan-1 (0.5408)        =28.41         =280                               

Explanation / Answer

Its just a conversion from km/hr to  m/s .    1 km/hr =     1000 m /3600 s   = 10 / 36   m/s    =    5/18  m/s . Notice in the problem they give you the speed in km/hr, butyou have to work with standard units of m/s. So the person whowrote the solution converted in the middle of the calculation bymultiplying by   5/18.

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