A student is about to do a push-up. Her

center of gravity lies directly above P on the floor, which is 1.2 m from her feet and 0.6 m from her hands.If her mass is 56 kg, what is the force exerted by the floor on her hands? The acceleration of gravity is 9.81 m/s2. Answer in units of N

The total moment about the toes is zero.

Therefore,

F*1.8 m - M*g*1.2 m = 0

Solve for F, the force on the hands

To find the force exerted by the floor on the student's hands, we need to consider the torque and gravitational forces acting on the student's body.

First, let's calculate the torque due to the student's weight. The torque is given by the product of the weight (mg) and the distance between the center of gravity and the pivot point. In this case, the pivot point is the hands, and the distance is 1.2 m since her center of gravity is directly above P on the floor.

Torque = weight * distance from pivot point
Torque = mg * 1.2 m

Next, let's calculate the weight of the student. The weight is given by the product of the mass (m) and the acceleration due to gravity (g). In this case, the mass is 56 kg and the acceleration due to gravity is 9.81 m/s².

Weight = mass * acceleration due to gravity
Weight = 56 kg * 9.81 m/s²

Now, let's substitute the weight into the torque equation:

Torque = (56 kg * 9.81 m/s²) * 1.2 m

The torque created by the student's weight produces a counterclockwise rotation around the pivot point (hands). To balance this torque and maintain equilibrium, the floor exerts an equal and opposite clockwise torque on the student's hand.

The torque exerted by the floor on the student's hands can be calculated as:

Torque = force * distance from pivot point

Since the torque due to the weight must be balanced by the torque exerted by the floor, we can equate the two and solve for the force:

(56 kg * 9.81 m/s²) * 1.2 m = force * 0.6 m

Now, let's solve for the force:

force = (56 kg * 9.81 m/s² * 1.2 m) / 0.6 m

Finally, substitute the values into the equation to calculate the force:

force = (66.192 N * 1.2 m) / 0.6 m
force = 132.384 N

Therefore, the force exerted by the floor on the student's hands is 132.384 N.