Where is the center of gravity of this person

[Gauss177]

1. A 160 cm tall person lies on a light (massless) board which is supported by two scales, one under the feet and one beneath the top of the head. The scale under the feet reads 29.4 kg and the one under the head reads 32.8 kg. Where is the center of gravity of this person measured from the bottom of the feet?

My thoughts:
The person would feel 321.4 N (32.8 kg * 9.8) pushing up at the head, and 288.1 N (29.4 kg * 9.8) pushing up at the feet. The force of mg would be an arrow pointing down under the center of gravity, located x meters from the bottom of the feet. For person to be in equilibrium, Fnet must equal 0, so:

321.4 N + 288.1 N = F
F = 609.6 N

Also, torque must also equal 0, so:

321.44(1.6) = 609.56(x)
x = .844 m, or 84.4 cm from the bottom of the feet.

Is this correct?

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2. A traffic light hangs from a structure as shown in the attached picture. The uniform aluminum pole AB is 4.5 m long and weighs 5.0 kg. The weight of the traffic light is 10.0 kg. Determine the tension in the horizontal massless cable CD, and the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole.

My thoughts:
I drew a free body diagram of the aluminum pole AB, and made the point A as my "fulcrum." Then, I would only have to worry about mg of the pole (5*9.8=49 N), Mg of the light (10*9.8=98 N), and tension T exerted by the cable CD. I split T into its components, Tx and Ty, and since Fnet must equal 0 for it to be at equilibrium:

Ty = 49 + 98 = 147 N
Therefore, Tx = 147 tan 37 = 110.72 N

I'm really not sure how to do the second part: finding the vertical and horizontal components of the force exerted by the pivot on the pole. I'm probably not visualizing it correctly, so please help.

 

[Andrew Mason]

Also, torque must also equal 0, so:

321.44(1.6) = 609.56(x)
x = .844 m, or 84.4 cm from the bottom of the feet.

Is this correct?

The torque about any point must be 0. Let's pick the centre of mass (you can pick either fulcrum and work out the expression for the torque of the centre of mass about that point).

The torque about the centre of mass must be 0. Let's say the centre of mass is a distance x from the feet. The torque is:

$$\tau = F_{head}\times (L-x) + F_{feet}\times x = 0$$

Work that out.

AM

 

[m2003]

Your calculations for the first part appear to be correct. The center of gravity of the person would be 84.4 cm from the bottom of their feet.

For the second part, to find the vertical and horizontal components of the force exerted by the pivot on the pole, you can use the concept of torque again. Since the pole is in equilibrium, the net torque around point A must be zero. This means that the torque exerted by the pivot must be equal and opposite to the torque exerted by the weight of the pole and the traffic light.

To find the torque exerted by the weight of the pole and traffic light, you can use the equation torque = force x distance. The distance is the length of the pole, 4.5 m. The force is the weight, which is 5 kg for the pole and 10 kg for the traffic light. So the torque exerted by the weight of the pole is 5 kg * 9.8 m/s^2 * 4.5 m = 220.5 Nm. The torque exerted by the weight of the traffic light is 10 kg * 9.8 m/s^2 * 4.5 m = 441 Nm.

Since the net torque around point A must be zero, the torque exerted by the pivot must also be 220.5 Nm + 441 Nm = 661.5 Nm.

Now, the force exerted by the pivot on the pole has both a vertical and a horizontal component. The vertical component is equal to the weight of the pole and traffic light, which is 5 kg * 9.8 m/s^2 + 10 kg * 9.8 m/s^2 = 147 N.

To find the horizontal component, you can use the equation torque = force x distance again, but this time you will use the distance from point A to the point where the force is applied. In this case, that distance is the length of the pole, 4.5 m. So the horizontal component of the force exerted by the pivot on the pole is 661.5 Nm / 4.5 m = 147 N.

Therefore, the force exerted by the pivot on the pole has a vertical component of 147 N and a horizontal component of 147 N.