Question

John ( body mass= 65 kg) is taking off for a long jump . Horizontal accerleration ax is 5m/s^2 and vertical acceleration ay is 0.9 m/s^2. a) calculate the horizontal ground reaction force F g,x .
b) calculate the vertical ground reaction force Fg,y.
c) calculate the resultant ground reaction force Fg?

Answer 1

Answer:

(a) The horizontal ground reaction force  $F_{g,x}=325\, N$

(b)  The vertical ground reaction force $F_{g,y}=696\, N$

(c)   The resultant ground reaction force $F_g=768\, N$

Explanation:  

Given

John mass , m = 65 kg

Horizontal acceleration , $a_x= 5.0 \frac{m}{s^{2}}$

Vertical acceleration , $a_y=0.9 \frac{m}{s^{2}}$

(a) Using Newton's 2nd law in horizontal direction

$F_{g,x}=ma_x$

=>$F_{g,x}=65\times 5\, N=325\, N$

Thus the horizontal ground reaction force  $F_{g,x}=325\, N$

(b) Using Newton's 2nd law in vertical direction

$F_{g,y}-mg=ma_y$

=>$F_{g,y}=mg+ma_y$

=>$F_{g,y}=65\times (9.81+0.9)\, N=696\, N$

Thus the vertical ground reaction force $F_{g,y}=696\, N$

(c)  Resultant ground reaction force is

$F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}$

=>$F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N$

=>$F_g=768\, N$

Thus  the resultant ground reaction force $F_g=768\, N$