A woman weighing 580n does a pushup from her knees. What are the normal forces of the floor on each of her hands and
Answers
Answer:
Given data:
The weight of the women is,
W=580N
If the reaction forces on hands and knees are,
AH and BK .
The condition here can be considered as the condition of point loading on a beam resting with two hinged supports,
Schematic diagram
Thus, by the expression of the force equilibrium in vertical direction,
AH+BK=580 ---------------- (1)
By taking the moment of forces about point B is,
580×54−AH×76=0AH
=
412.10
N
Substituting this value in equation (1),
412.10
+
B
K
=
580
B
K
=
580
−
412.10
B
K
=
167.9
N
The forces at hand position will get distributed equally to both hands, thus,
A
H
2
=
412.10
2
=
206.05
N
Thus the force on each hand is,
206.05
N
And,
The forces at knees position will get distributed equally to both knees, thus,
B
K
2
=
167.9
2
=
83.95
N
Thus the force on each knee is,
83.95
N
Answer:
force on each knee is
83.95