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

(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 (m)/(s^(2))`

Vertical acceleration ,
`a_y=0.9 (m)/(s^(2))`

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

`F_(g,x)=ma_x`

=>
`F_(g,x)=65* 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* (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))^{(1)/(2)}`

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

=>
`F_g=768\, N`

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