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Alika [10]
3 years ago
12

A man is doing push-ups. he has a mass of 68 kg and his center of gravity is located at a horizontal distance of 0.70 m from his

palms and 1.00 m from his feet. find the forces exerted by the floor on his palms and feet.
Physics
2 answers:
LUCKY_DIMON [66]3 years ago
8 0

Answer:

F_{palms}=392N

F_{feet}=274N}

Explanation:

To solve this problem we need to be aware that the man is rotational and mechanical equilibrium, it means that the sum of all torques for a fixed rotation axis is zero and the net force on him is zero too. Since we are working in a two-dimensional space we can work with the magnitude of the torque and considering upwards as positive.

\tau=r*R*sin(\theta),

where F is the force applied, r*sin(\theta) is the perpendicular distance from the axis of rotation to the line of action of the force. In this particular case, all forces are perpendicular to this distance so our expression reduces to

\tau=r*R.

Now we can start solvig our problem. We are going to use

Using the feet as the axis of rotation

\sum \tau=\tau_{feet}+\tau_{palms}-\tau_{weight}=0,

\sum \tau=0*F_{feet}+1.7+F_{palms}-1*mg=0,

\tau=0*F_{feet}+1.7+F_{palms}-1*mg=0,

solving for F_{palms}

F_{palms}=\frac{mg}{1.7}=\frac{68*9.8}{1.7}

F_{palms}=392N.

To find the force exerted on his feet we use Newton's second law

\sum F=ma,

since we are in mechanical equilibrium

\sum F=0,

so

F_{palms}+F_{feet}-mg=0

F_{feet}=mg-F_{palms}

F_{feet}=274N}.

lidiya [134]3 years ago
7 0
Mass m = 68 kg
center of gravity from his palms x = 0.7 m
center of gravity from his feet x ' = 1 m
forces exerted by the floor on his palms and feet are F and F ' respectively.

with respect to palms :---------------------

( F*0 ) - (W * x ) + [ F ' * (x+x') ] = 0

             -mg*0.7 + F ' * 1.7 = 0    where W = weight = mg

F ' * 1.7 = mg * 0.7

          F ' = mg * 0.7 / 1.7

               = 68 *9.8 * ( 0.7 / 1.7 )

               = 274.4 N

with respect to feet :--------------------

( F ' * 0 ) -( W* x ' ) + [F * ( x + x') ] = 0

                -mg*1 + [ F * 1.7 ]= 0

                                 F = mg / 1.7

                                    = 392 N
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the position vector (x,y) will be (1.5 m,-2.25 m) and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s) when x reaches its maximum x coordinate

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when x reaches its maximum coordinate , then vx=0

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vx²=v₀x²+2*ax*x

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vx = v₀x + ax*t → t= (vx-v₀x)/ax = (0- 3 m/s)/  (−1.00 m/s²) = 3 seconds

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y = y₀+v₀y*t + 1/2 ay*t²

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v₀y = initial velocity in the y direction = 0 m/s

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y= coordinates in the y-axis

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then since y₀=0 and v₀y=0

y = 1/2*ay*t²

y = 1/2*ay*t² = 1/2*(−0.5 m/s²)*(3 s)² = -2.25 m

and

vy=v₀y+ ay*t= 0+(−0.5 m/s²)*(3 s)= (-1.5 m/s)

therefore the position vector (x,y) will be (1.5 m,-2.25 m)

and the velocity vector (vx,vy) will be ( 0 m/s , -1.5 m/s)

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