```Interactive Free-Body Diagram (FBD): StaticsA point mass m is on a ramp inclined by angle  we wish to maintain static equilibrium for this point mass on the ramp by applying a force F inclined by angle , as shown in the diagram below. The static coefficient of friction between the point mass and ramp is S.g Y XFmWith the point mass assumption, there can be no box rotation by definition. The statics freebody diagram (FBD) for the point mass box is shown in the figure below:FW FfNIn the XY coordinate system shown (X is aligned with the ramp), these four force vectors are expressed as: WWsin     cos  F  F cos      F sin     N  0  N Ff  Ff  0  The primary purpose of this exercise is to allow the user to change various system parameters, calculate the required force F for static equilibrium, see the effects on the FBD, and feel each of the F  0 . It is forces. From the free-body diagram, we apply the vector equation of static equilibrium, convenient to use the XY coordinate system shown in the first diagram above: X is along the ramp F  0 yields: direction and Y is normal to the ramp.X:F Y :Fx y F cos     F fW sin  0N  F sin      W cos0 S N , upwards along the ramp to resist motion down the ramp, as shown in the free-body diagram; N is the normal force of the ramp acting on the point mass. The weight force is W  mg , trying to move the point mass down the ramp. Substituting these given parameters into the statics equations yields:The static friction force is F f F cos      S N F sin      N mg sin  mg cosThese are two linear equations coupled in the two unknowns F, N. It is convenient to solve this using a matrix approach (high-school algebra is fine too). The matrix-vector expression of the above equations is:  cos     S   F   mg sin        sin     1   N  mg cos
The matrix-vector solution for the unknowns is: F  cos     S    N   sin     1
1mg sin   1  1  S   mg sin        cos    mg cos     sin     mg cos
Where cos      S sin     is the matrix determinant. The solution is:Fmg sin    S cos   cos      S sin    Nmg cos    cos  sin     sin   cos      S sin    Using the cosine sum-of-angles formula from trigonometry ( cos a cos b  sin a sin b  cosa  b  ), the solution for N simplifies: Nmg cos  cos      S sin    To satisfy static equilibrium, the above expression for F must be applied; otherwise dynamics motion will result. An interesting side problem is to apply zero force F and calculate the maximum ramp angle  for static equilibrium for a given m and S. In this case the statics equations are:S N N mg sin  mg cosThe Y equation yields the unknown normal force: N yields the maximum angle  for static equilibrium:mg cos ; substituting this into the X equation MAXtan 1  SNote this result is independent of mass m. For this result, F = 0 will yield static equilibrium for any    MAX . 0  m  10 0   1 0  0 Computer sets: Visualize: equilibrium. User sets:m,  S , , and 90 90g = 9.81 m/s2, (down, not in the ­Y direction unless =0). Free-body diagram with forces to scale. In the alternate problem,  MAX for staticNumerical Display: F, N, Ff, and W; plus  MAX for alternate problem. User Feels: Forces F, N, Ff, or W (user chooses). Joystick should display the vector forces to the user's hand, either in horizontal/vertical or XY coordinates. Example: When the user enters m = 80 kg,   35 ,   50 , and   0.15 , the force for static equilibrium is F = 381.52 N and the associated normal force, friction force, and weight are N = 544.12 N, Ff = 81.62 N, and W = 784.80 N. For this coefficient of static friction, the maximum angle for static equilibrium with F = 0 is  MAX  8.53 .Comprehension Assignment: Once you get the `feel' for this simulation, run the program several times to collect and plot data: for m = 10 kg and a fixed value of the static coefficient of friction S, vary the ramp angle  and force angle  over their allowable ranges and determine the resulting forces F, N, Ff, and W. Plot F, N, Ff, and W vs. . For each case, there will be different curves for different  values. Repeat these plots for various values of S over its allowable range. For the alternate problem with F = 0, plot  MAX vs. S. Discuss the trends you see in all cases ­ do the results make sense physically?```

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