A fighter jet landing on an aircraft is made to brake very quickly by means of c

Question

A fighter jet landing on an aircraft is made to brake very quickly by means of cable that the plane connects to via a hook.   Assume that a 21,000 lb jet touches down at 170 mph and its speed is reduced to zero in 240 ft  Assuming constant deceleration, calculate the force applied to the jet by the braking cable.  In order to get a feel for the magnitude of he force, calculate the braking force acting on a 3500 lb car in order to bring it (at constant deceleration) from 60 mph to zero in 140 ft.  How many of these "car force units needed to obtain the force actin on the jet? 

Explanation / Answer

    a.     Initialspeed ofjet   uj   =   170   mph                                        =   170* 5280 / 3600   ft/s                                        =   249.33   ft/s    Final speed ofjet   vj   =   0   ft/s    Distancecovered   dj   =   240   ft    Mass ofjet   mj   =   21000   lb    Third equation of motion is    vj2   =   uj2   +   2* aj * dj    02   =   249.332   +   2* aj * 240   de acceleration   aj   =   -6.217 * 104 / 480                                  =   -129.51   ft/s2    -ve sign here indicated deaccleration and itcan be discarded.    Force   Fj   =   mj* aj                      =   21000* 129.51                      =   2.72* 106   lb - ft/s2
                                       =   60* 5280 / 3600   ft/s                                        =   88.0   ft/s    Final speed ofcar   vc     =   0   ft/s    Distancecovered   dc     =   140   ft    Massof car   mc     =   3500   lb    Third equation of motion is    vc2   =   uc2   +   2* ac * dc    02   =   882   +   2* ac * 140   deacceleration   ac   =   - 7744/ 280                                  =   - 27.657   ft/s2    -ve sign here indicated deaccleration and itcan be discarded.    Force   Fc   =   mc* ac                      =   3500* 27.657                      =   9.68* 104   lb - ft/s2                      =   3500* 27.657                      =   9.68* 104   lb - ft/s2    c.   Ratio offorces   Fj /Fc   =   2.72 *106 / 9.68 * 104                                                 =   28.099          i.e.    28.099   carforce units are required to obtain the braking force working on thejet.                                                 =   28.099          i.e.    28.099   carforce units are required to obtain the braking force working on thejet.

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