Question

A person stands on a scale on the ball of one foot at an angle of 13.3

Answer 1

a) Force= F=mgcos $\theta$ =67.5(10)cos13.3                     " mgsin $\theta$ will be the component of gravity whcin is balanced by

           =656.8N                                                                            frictional force."

b)     F= 67.5(10)cos30=584N                 The force is changed by 72.8N

c)      The normal force on one foot will be equal to = mgcos $\theta$ =656.8N for part a) and 584N for part b)

                  For two feet Newtons second law of motion could be written

                    mgcos $\theta$ -Fn-Fn=m(0)

                    mgcos $\theta$ =2Fn

                      Fn= mgcos $\theta$ / 2

                        Fn= 686.8/2

                            =343.4N

                         This will be the normal force when he is standing on both feets.

In case of 30 degree angle put the value of cos30 in above formula.