A 70 kg person is having a tug-of-war. The rope is horizontal and 1.3 m above the

ground. This person’s center of mass is 1.0 m from the ground when he is in an
upright standing position. If the force his opponent exerts on the rope is 400 N,
what angle must the person be from the horizontal to be in equilibrium? (Assume
the knees and hips are not bent).

write the sum of moments about his feet.

assume angle theta ..

400*1.3*sinTheta=70g*1.0*cosTheta

tan Theta=70(9.8)/400*1.3

theta= you do it.

I don't get this question still! please help

I don't understand how you got that solution. Could someone please elaborate? PLEASE

To solve this problem, we will use the principle of torque equilibrium. Torque is the rotational equivalent of force and is calculated by multiplying the force applied to an object by the distance from the axis of rotation.

In this case, the person is in equilibrium when the torque exerted by the opponent's force is balanced by the torque exerted by the person's weight.

First, let's calculate the torque exerted by the opponent's force. The torque exerted by a force is calculated by multiplying the magnitude of the force by the perpendicular distance from the axis of rotation to the line of action of the force.

Given:
Force exerted by the opponent: F = 400 N

The perpendicular distance from the axis of rotation (the person's center of mass) to the line of action of the force is the vertical distance from the person's center of mass to the ground, which is 1.0 m.

So, the torque exerted by the opponent's force is given by:
Torque_opponent = F * d_opponent = 400 N * 1.0 m

Next, let's calculate the torque exerted by the person's weight. The weight of an object is calculated by multiplying its mass by the acceleration due to gravity (9.8 m/s^2), and the torque exerted by the weight is the weight multiplied by the perpendicular distance from the axis of rotation to the line of action of the weight.

Given:
Mass of the person: m = 70 kg
Acceleration due to gravity: g = 9.8 m/s^2
Vertical distance from person's center of mass to the ground: 1.0 m

The weight of the person is given by:
Weight = m * g = 70 kg * 9.8 m/s^2

The perpendicular distance from the axis of rotation (the person's center of mass) to the line of action of the weight is the distance between the person's center of mass and the rope, which is given as 1.3 m.

So, the torque exerted by the person's weight is given by:
Torque_person = Weight * d_person = (70 kg * 9.8 m/s^2) * 1.3 m

To be in equilibrium, the torque exerted by the opponent's force must be equal to the torque exerted by the person's weight. Therefore, we can set up the following equation:

Torque_opponent = Torque_person

Simplifying and substituting the values we calculated:

400 N * 1.0 m = (70 kg * 9.8 m/s^2) * 1.3 m

Now, we can solve this equation to find the angle at which the person must be from the horizontal to be in equilibrium.