The first condition for equilibrium: The second condition for equilibrium: • • ΣF = 0 ΣΓ = 0
In when both of these conditions are satisfied in static systems all forces and torques sum to zero. In problems where the first and second conditions of equilibrium are satisfied, the best strategy is to create FBD’s for both the first and second conditions, derive equations based on these FBD’s and then see what useful information may be gleaned from these equations. When applying the second condition we are free to choose any axis about which to compute torques. It is best to choose an axis that eliminates one or more forces that have lines of force that pass through it.
Example 1 Consider a playground seesaw. The mass of the plank is 2.0 kg, the masses of two children on it are 25 kg and 30 kg with the 30 kg child sitting 2.5 meters from the center of the plank (the fulcrum) as shown below. Where must the second child sit in order for this system to be in equilibrium?
Noting that a normal force directed upwards acts at the point of the fulcrum, the FBD’s for the first condition yield:
= N − mc1 g − m p g − mc 2 g = 0 →
= N − 294 N − 19.6 N − 245 N = 0
Note that while this is all true it is not, by itself, particularly useful. To apply the second condition we must first choose an axis about which to compute torques. The axis that makes the most physical sense would be one directly through the board over the fulcrum, but we could choose any axis that made computations easier. In this case choosing the axis associated with the fulcrum eliminates the forces created by the mass of the board itself since these act on the center of mass of the board which is located directly over the fulcrum.
The FBD’s for the second condition yield:
∑ Γ = (294 N )(2.5meters) − (245 N )( xmeters) = 0
Solving this equation for x yields a distance of 3 meters.
Example 2 Consider the following cantilevered beam:
The beam has a mass of m = 25 kg and is 2.2 meters long. The suspended block has a mass M = 280 kg and the supporting cable makes an angle of 300 with the beam. Determine the force that the wall exerts on the beam at the hinge and determine the tension in the supporting cable.
Notice that the normal force is the x component of the force exerted by the wall on the beam through the hinge (Fx). Because the beam is also held up by the hinge (Fy) the total force the wall exerts on the beam is the aggregate of these two components. So we must determine, from the available information, Fx, Fy, Tx, Ty and finally T and F.
Application of the first condition with our sign convention yields:
= Fx − Tx = 0 ∴ Fx = Tx = Fy + T y − mg − Mg = 0
Application of the second condition with respect to the hinge yields:
Fy – line of action passes through the hinge → no torque Fx – line of action passes through the hinge → no torque Tx – line of action passes through the hinge → no torque mg – exerts a torque Mg – exerts a torque Ty – exerts a torque With our sign convention:
∑ Γ = −(mg )(1.1m) − (Mg )(2.2m) + (T
)(2.2m) = 0 )(2.2m) = 0
∑ Γ = −(245 N )(1.1m) − (2744 N )(2.2m) + (T
∑ Γ = −270 N ⋅ m − 6037 N ⋅ m + (T
)(2.2m) = 0 → T y = 2867 N
Since T y = T sin θ : T = 5734 N , and with a little more work Tx = 4966 N . With the magnitude of T and all of its components known, it is a simple matter to substitute into the equation in y from the first condition and solve for Fy:
= Fy + 2867 N − 245 N − 2744 N = 0
Fy = 122 N
Noting that Fx = Tx (why?), Fx = 4966N → F =
(4966 N ) 2 + (122 N ) 2 = 4967 N
tan o = /
y 122 N = ∴ o = 1.4° / x 4966 N
v F = 4967 N @ 1.4°
Example 3. A 5 meter long ladder leans against a frictionless wall. The point of contact between the ladder and the wall is 4 meters above the ground. The ladder is uniform with a mass of 12 kg....