A boy carries a sack on one end of a very light stick that is balanced on his shoulder, at an angle of 22.0 degrees up from the horizontal. The mass of the sack is 7.00 kg, and it sits 1.20 m from his shoulder. If his hand is 0.350 m from his shoulder, on the other end of the stick, what is the magnitude of the downward force the boy exerts with his hand?

To find the magnitude of the downward force the boy exerts with his hand, we can first calculate the torque exerted by the sack and then determine the force required to balance the torque.

Torque is defined as the product of force and the perpendicular distance from the point of rotation (in this case, the shoulder). The torque exerted by the sack can be calculated using the following formula:

Torque = (force of the sack) * (perpendicular distance from the shoulder to the sack)

The force of the sack can be determined using the formula:

Force of the sack = (mass of the sack) * (gravitational acceleration)

The perpendicular distance from the shoulder to the sack can be calculated using trigonometry:

Perpendicular distance = (distance from the shoulder to the sack) * (sin(angle))

Let's substitute the given values into the equations:

Distance from the shoulder to the sack = 1.20 m
Distance from the shoulder to the hand = 0.350 m
Angle = 22.0 degrees
Mass of the sack = 7.00 kg
Gravitational acceleration = 9.8 m/s^2

First, let's calculate the perpendicular distance from the shoulder to the sack:

Perpendicular distance = (1.20 m) * sin(22.0 degrees) ≈ 0.45 m

Next, let's calculate the force of the sack using the formula:

Force of the sack = (7.00 kg) * (9.8 m/s^2) ≈ 68.6 N

Finally, we can calculate the torque exerted by the sack:

Torque = (68.6 N) * (0.45 m) ≈ 30.9 N·m

To balance the torque, the boy needs to exert an equal and opposite torque with his hand. Since the horizontal distance from the shoulder to the hand is given as 0.350 m, we can calculate the force required using the formula:

Force required = Torque / (perpendicular distance from the shoulder to the hand)

Force required = (30.9 N·m) / (0.350 m) ≈ 88.3 N

Therefore, the magnitude of the downward force the boy exerts with his hand is approximately 88.3 N.