Static Equilibrium and downward force

[Meowzers]

I just want to make sure that I approached the following problem correctly:

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?

I drew somewhat of a force diagram, then:

Set x=force of hand
torque of hand + torque of wall = 0
rFcos(theta) + rFsin(theta) = 0
-(.350m)(x)(cos 22) + (1.2m)(9kg*9.8 m/s2)(sin 22) = 0
.3245x = 39.648
x = 122.18 N

 

[PhanthomJay]

Besides a couple of typos, why did you use the sin theta distance component as the moment arm for the downward force of the boy's hand? And oh if you use weight on right side of equation, you need to use it on left side also (i.e., multiply the mass by g.)

 

[Meowzers]

Well, in my force diagram, the 22 degree angle is right across from the downward force of the boy's hand. I guess they're just mirror images of each other...

So, the force of both the hand and the sack are Fsin(theta)?

torque of hand + torque of wall = 0
rFsin(theta) + rFsin(theta) = 0
-(.350m)(x)(sin 22) + (1.2m)(9kg*9.8 m/s2)(sin 22) = 0
.13111x = 39.648
x=302.4 N

 

[PhanthomJay]

Well, in my force diagram, the 22 degree angle is right across from the downward force of the boy's hand. I guess they're just mirror images of each other...

So, the force of both the hand and the sack are Fsin(theta)?

torque of hand + torque of wallyou mean sack? = 0
rFsin(theta) + rFsin(theta) = 0
-(.350m)(9.8)x(sin 22) + (1.2m)(9 or 7??kg*9.8 m/s2)(sin 22) = 0
.13111x = 39.648
x=302.4 N

Without a diagram, it is hard to understand where the 22 degree angle is; but in your statement you said it was up from the horizontal, in whichcase you should use cos 22 on both sides of the equation. There are 2 ways to calculate torques: The first is to take the magnitude of the force and multiply it by the perpendicular distance from the line of action of the force to the point you are taking moments about. In which case you have, for the sack torque,
$$F(r) cos\theta = 9(9.8)(1.2) cos 22$$.
OR, you can use the definition of torque
$$T =Frsin\theta$$, but here, $$\theta$$is the included angle between the force and displacemnt vectors, $$\theta = 68 degrees$$.. You will get the same answer; I prefer the first method.

 

[Meowzers]

Thanks for all your help :)