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There are many structural problems in statics that do not involve a truss.  For example, a beam is not loaded axially.  A beam is analyzed in terms of bending moments and shear.  Many simple tools can be analyzed with statics.

==Example 1==

===Question===

[[Image:StaticsPliers1.svg|thumb]]

The picture to the right shows a tool used to compress springs.  In the tool's current state a person is squeezing the pliers with 40 Newtons applied to each handle to hold the spring(shown in orange) in a steady partially compressed state.  The black circle is a pin (a pin can transfer forces, but cannot resist turning).  The forces used to squeeze the pliers are vertical in addition to the forces applied on the spring.

Find the compressive force on the spring and the force on the pin.

====Hint====
Think of how the plier components can be disassembled and analyzed for equilibrium.



===Answer===

[[Image:StaticsPliersIsolated.svg|thumb|left]]

If the pin in the question's picture was welded to the hole it fits in, the force from the spring could be any value; however, the pin is free to rotate (it can't resist a turning moment) so the picture to the left can be used for further analysis.

If moments are summed about the black pin (let's designate it "pin A"), the spring force(designated "F") can be solved for.

<math>\sum M_A = 0 = -6 in*40N + 2 in*F</math>

<math>\ F = 120N </math>

This means the spring is being compressed with 120N.

The force in the pin can be solved for with force equilibrium.  Let's call the pin's <math> x </math> force <math> P_x </math> and the <math> y </math> force <math> P_y </math>.

<math>\sum F_x = 0 = P_x </math>

<math>\sum F_y = 0 = 40N + P_y + 120N </math>

<math>\ P_y = -160N </math>

Since the equations were written in terms of the assumed force directions pictured and since <math> P_y </math> is negative, the force acts in the opposite direction as shown in the picture.

[[Image:StaticsPliersPin.svg|thumb]]

'''The pin exerts a downward force on the handle shown in the left picture, but it exerts an upward force of the same magnitude on the other handle.'''  This statement is very important and can be further explained by the picture to the right.  The picture shows a dark gray and light gray rectangle.  The light and dark gray corresponds to the previous pictures.  The black rectangle represents the pin which goes through both handles as indicated with dashed lines.  The dark handle is exerting an upward force of 160N on the pin and the light handle is exerting a downward force of 160N on the pin.  Because the forces are equal and opposite the pin does not travel vertically.  The force from the dark handle is transmitted through the pin to the light handle and the force from the light handle is transmitted through the pin to the dark handle.  '''The pin is in 160N of shear'''.

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