Using Mathcad for Statics and DynamicsBedford and Fowler2.4 - 2.62.3 - 2.52. Dot Products2.92.53. Equilibrium of a Particle, Free-Body Diagrams3.33.2 - 3.34.2 - 4.32.6, 4.35. Moment of a Couple4.64.46. Reduction of a Simple Distributed Loading4.107.35.1 - 5.35.1 - 5.28. Dry Friction8.29.19.
![Force System Mathcad Force System Mathcad](/uploads/1/2/5/4/125448262/846340905.png)
Set unit system to US type force:1lbf type force= Mathcad returns 32.174 lb.ft/sec^2 you can then tab to the units placeholder and type lbf and Mathcad will display force=1lbf, indicating that lbf is indeed defined as a unit, which is also evident from the Insert Unit selections. Lbm-sec system, the force unit is determined by length (in.), mass. The addition of the custom default unit system in Mathcad version 13 is a great improvement.
Finding the Centroid of Volume9.27.49.57.410.1 - 10.28.1 - 8.212. Curve-Fitting to Relate s-t, v-t, and a-t Graphs12.32.213. Curvilinear Motion: Rectangular Components12.52.314. Curvilinear Motion: Motion of a Projectile12.62.315. Curvilinear Motion: Normal and Tangential Components12.72.316.
![Force System Mathcad Force System Mathcad](/uploads/1/2/5/4/125448262/338870502.png)
Dependent Motion of Two Particles12.9ch. Kinetics of a Particle: Force and Acceleration13.43.1 - 3.418. Equations of Motion: Normal and Tangential Components13.53.419. Principle of Work and Energy14.38.1 - 8.220. Rotation About a Fixed Axis16.39.14. Cross Products and Moments of Forces7. Equilibrium of a Rigid Body10.
Resultant of a Generalized Distributed Loading 11. Calculating Moments of InertiaDynamics1. Resolving Forces, Calculating ResultantsStaticsHibbelerby Ron Larsen and Steve HuntRef: Hibbeler § 2.4-2.6, Bedford & Fowler: Statics § 2.3-2.5Resolving forces refers to the process of finding two or more forces which, when combined, will produce a force with the same magnitude and direction as the original. The most common use of the process is finding the components of the original force in the Cartesian coordinate directions: x, y, and z. A resultant force is the force (magnitude and direction) obtained when two or more forces are combined (i.e., added as vectors).Breaking down a force into its Cartesian coordinate components (e.g., Fx, Fy) and using Cartesian components to determine the force and direction of a resultant force are common tasks when solving statics problems. These will be demonstrated here using a two-dimensional problem involving coplanar forces. Example: Co-Planar Forces Two boys are playing by pulling on ropes connected to a hook in a rafter.
The bigger one pulls on the rope with a force of 270 N (about 60 lbf) at an angle of 55° from horizontal. The smaller boy pulls with a force of 180 N (about 40 lbf) at an angle of 110° from horizontal. Which boy is exerting the greatest vertical force (downward) on the hook? What is the net force (magnitude and direction) on the hook – that is, calculate the resultant force.-55°N-110°Note: The angles in this figure have been indicated as coordinate direction angles. That is, each angle has been measured from the positive x axis.271 801Resolving Forces, Calculating Resultants0NSolution First, consider the 270 N force acting at 55° from horizontal.
The x- and y-components of force are indicated schematically, asFx 55°Fy27 0NThe x- and y-components of the first force (270 N) can be calculated using a little trigonometry involving the included angle, 55°: cos(55°) =Fx1, or 270 NFx1 = (270 N ) cos(55°)and sin(55°) =Fy1 270 N, orFy1 = (270 N ) sin(55°).Mathcad can be used to solve for Fx1 and Fy1 using its built-in sin and cos functions, but these functions assume that the angle will be expressed as radians, not degrees. You can use the deg unit to explicitly tell Mathcad that the angle is in degrees and must be converted (by Mathcad) to radians. Fx1:= ( 270⋅ N) ⋅ cos ( 55⋅ deg ) Fx1 = 154.866N Fy1:= ( 270⋅ N) ⋅ sin ( 55⋅ deg ) Fy1 = 221.171NYour Turn180N110°Answer, part a) The larger boy exerts the greatest vertical force (221 N) on the hook.
The vertical force exerted by the smaller boy is only 169 N.Fx20°FyShow that the x- and y-components of the second force (180 N acting at 110° from the x-axis) are 61.5 N (-x direction) and 169 N (-y direction), respectively. Note that trigonometry relationships are based on the included angle of the triangle (20°, as shown at the right), not the coordinate angle (-110° from the x-axis).Solution, continued To determine the combined force on the hook, FR, first add the two y-components calculated above, to determine the combined y-directed force, FRy, on the hook: FRx0N221 N27180 N77°FRyFR169 NFRy:= Fy1 + Fy2 FRy = 390.316NThe y-component of the resultant force is 390 N (directed down, or in the –y direction). Note that the direction has not been accounted for in this calculation. Then add the two x-components to determine the combined x-directed force, FRx, on the hook. Note that the two x-component forces are acting in opposite directions, so the combined x-directed force, FRx, is smaller than either of the components, and directed in the +x direction.FRx27180 N77°0N62 N FRyFR155 NFRx:= Fx1 + −Fx2FRx = 93.302NThe minus sign was included before Fx2 because it is directed in the –x direction. The result is an xcomponent of the resultant force of 93 N in the +x direction.