When forces are applied to rigid bodies, the conditions of equilibrium can be used to determine any unknown forces, and the reactions at the supports.

Example 1: Spring Support

A force, F, is applied to the end of a bar to stretch a spring until the bar is horizontal. If the spring constant is k = 600 N/m and the unstretched length of the spring assembly (spring and connecting links) is 75 mm, and ignoring the mass of the bar, determine:

a)the magnitude of the applied force, F, and

b)the reactions at point A.

Solution

First, a free-body diagram is drawn.

We have assumed that the reactions at A will include both x and y components, and will calculate the actual values of Ax and Ay as part of the solution process.

First, a little trigonometry is required to obtain the extended length of the spring assembly.

120mm Lext := cos (28deg)

Lext = 136mm

Then, we can calculate the actual extension of the spring…

Lunstretched := 75mm

Lstretch := Lext − Lunstretched

Lstretch = 61mm

…and the spring force, Fspring.

N kspring := 600 m

Fspring := kspring Lstretch

Fspring = 36.5N

Next, we calculate the x and y components of the spring force, using the angle from the positive x axis (not just 28°) so that the direction of the force is accounted for.

At equilibrium the sum of the moments at A must be zero. We can use this to solve for the applied force, F.

Fsp_x (16mm) + Fsp_y (120mm) + F(150mm) 0

Solve for F...

F := − Fsp_x (16mm) + Fsp_y (120mm) 150mm

Notes:

The first equation in the last example block states the equilibrium relationship, but was not actually solved by Mathcad. Only the equation after “Solve for F…” was actually used to calculate F.

The absolute value operator was used on Fsp_x since the direction of rotation was accounted for as the equilibrium equation was written. Here, counter-clockwise rotation was assumed positive.

The equilibrium relationships for the x and y components of force can be used to determine the reactions at A. First, x-component equilibrium requires that the sum of the x components of force be zero. This relationship is used to solve for Ax.

Ax + Fsp_x 0

Solve for Ax.

Ax := −Fsp_x

Finally, the equilibrium relationship for the sum of the y components of force is used to calculate Ay.

Ay + Fsp_y + F 0

Solve for Ay.

Ay := −Fsp_y − F

Ay = 0.0105N

Annotated Mathcad Worksheet

Spring Support Problem

Find the total extended length of spring assembly.

120mm Lext := cos (28deg)

Lext = 136mm

Calculate the spring's extension.

Lunstretched := 75 mm

Lstretch := Lext − Lunstretched

Use the sum of the moments about A to find applied force F.

At equilibrium, the sum of the moments at A must be zero. Here, counter-clockwise rotation is considered positive.

Fsp_x (16 mm) + Fsp_y (120 mm) + F (150 mm) 0

Solve for F...

F := − Fsp_x (16 mm) + Fsp_y (120 mm)

150 mm

F = −17.2N acting to cause clockwise rotation

Use x and y component force balances to solve for the reactions at A.

At equilibrium, the sum of the x components of force must be zero.

At equilibrium, the sum of the y components of force must be zero.

Ay + Fsp_y + F 0

Solve for Ay.

Ay := −Fsp_y − F

Ay = 0.0105N