Question

John(body mass=160pounds) is taking off for a long jump. The average ground reaction force Fg at takeoff is 1400 N pointing forward at an angle of 35° above the horizontal. (a) Draw a free body diagram of the person, specifying ground reaction forces Fg,y and Fg,x.
(b)Calculate ground reaction forces F g,y and Fg,x.
(c)What is the acceleration ax?
(d) What is the acceleration ay?

Answer 1

Answer:

The free body diagram of John is shown in the attached figure (in the FBD john's mass is supposed to be concentrated at his center of mass and FBD is made of center of mass)

b) As shown in the FBD the ground reaction forces are:

i) In X direction $F_{x}=1400cos(35^{o})=1146.81N$

ii) In Y direction $F_{y}=1400sin(35^{o})=803.0N$

c) The respective accelerations in x and y direction's is calculated by newton's second law as indicated under

$\sum F_{x}=ma_{x}\\\\\therefore a_{x}=\frac{\sum F_{x}}{m}=\frac{1146.8N}{72.57kg}=15.80m/s^{2}\\\\\sum F_{y}=ma_{y}\\\\\therefore a_{y}=\frac{\sum F_{y}}{m}=\frac{803.00-72.57\times 9.81}{72.57}=1.255m/s^{2}$