How Does the Boy Balance the Forces While Carrying the Sack?

[lacar213]

Homework Statement

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?

Homework Equations

net torque = 0

The Attempt at a Solution

7.00 * 1.20 / .350 = 24 N

 

[LowlyPion]

Homework Statement

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?

Homework Equations

net torque = 0

The Attempt at a Solution

7.00 * 1.20 / .350 = 24 N

And you got from kg to N how?

 

[thomate1]

This solution is incorrect. In order to find the magnitude of the downward force the boy exerts with his hand, we must first find the net torque acting on the system. To do this, we need to consider the forces acting on the sack and the stick.

First, let's draw a free body diagram of the system:


From the free body diagram, we can see that there are three forces acting on the sack and the stick: the weight of the sack (mg), the weight of the stick (mgsin22.0), and the downward force exerted by the boy's hand (F).

Next, we need to set up an equation for the net torque acting on the system. Since the system is in static equilibrium, the net torque must be equal to 0. We can write this as:

Στ = 0

Where Στ represents the sum of all the torques acting on the system. We can calculate the torque of each force by using the equation τ = rFsinθ, where r is the distance from the pivot point (the boy's shoulder) to the point where the force is applied, F is the magnitude of the force, and θ is the angle between the force and the lever arm.

Using this equation, we can write the following equation for the net torque:

Στ = (1.20 m)(mg) + (1.55 m)(mgsin22.0) - (0.350 m)(F) = 0

We know the mass of the sack is 7.00 kg, so we can substitute this in for m. We also know that g = 9.8 m/s^2. Substituting these values in, we get:

(1.20 m)(7.00 kg)(9.8 m/s^2) + (1.55 m)(7.00 kg)(9.8 m/s^2sin22.0) - (0.350 m)(F) = 0

Simplifying this, we get:

83.16 + 24.55 - 0.350F = 0

Solving for F, we get:

F = 107.71 N

Therefore, the magnitude of the downward force the boy exerts with his hand is 107.71 N. This is the force