Newtonian Mechanics vector solution needed to find the range of forces (P) that satisfy Fnetx = 0

% Find vector of P values to keep block in equilibrium

% Variables Given
m = 100; % mass, kg
angle_ramp = 15; % degrees
angle_bar = 20; % degrees
g = 9.81; % gravity (m/s^2)
coeff = 0.3; % coefficient of static friction 

% Calculations
P = [-600:0.1:600]; % vector of range for P (Newtons)
W = m*g;
Px = P*cosd(angle_bar); Py = sind(angle_bar);
N = W*cosd(angle_ramp)-Py;
Fx =  W*sind(angle_ramp); Fs = coeff*N;

% Find P values for Fnet(x) = 0
if P > 0 % pulling force
    Fnetx = Px - Fx - Fs; 
else  % pushing force or zero force applied
    Fnetx = Px - Fx + Fs; 
end

find(Fnetx==0)

ans =

  1×0 empty double row vector

Answer 1

[Torsten]

Your search will never find a value for P for which Fnetx is exactly 0.

Better use "fsolve" to solve for the corresponding root of Fnetx(P) = 0.

P0 = 600;
P = fsolve(@fun,P0,optimset('Display','none'))

P = 516.3332

[res,Px,Py] = fun(P)

res = -7.9902e-08

Px = 485.1945

Py = 176.5964

P0 = -30;
P = fsolve(@fun,P0,optimset('Display','none'))

P = -36.2812

[res,Px,Py] = fun(P)

res = 3.4890e-10

Px = -34.0932

Py = -12.4089

function [Fnetx,Px,Py] = fun(P)
  % Variables Given
  m = 100; % mass, kg
  angle_ramp = 15; % degrees
  angle_bar = 20; % degrees
  g = 9.81; % gravity (m/s^2)
  coeff = 0.3; % coefficient of static friction 
  % Calculations
  W = m*g;
  Px = P*cosd(angle_bar); Py = P*sind(angle_bar);
  N = W*cosd(angle_ramp)-Py;
  Fx =  W*sind(angle_ramp); Fs = coeff*N;
  % Find P values for Fnet(x) = 0
  if P > 0 % pulling force
    Fnetx = Px - Fx - Fs; 
  else  % pushing force or zero force applied
    Fnetx = Px - Fx + Fs; 
  end
end

Or use

[~,idx] = min(Fnetx.^2);
P(idx)

at the end of your code to get an approximate value for P.

[Doug Leaffer]

It's a close approximation @Torsten

Answer 2

[Doug Leaffer]

Thank you @Torsten

I found another MATLAB code solution which is accurate as I checked it in Excel using Goal Seek

=> Block remains in equilibrium for -36.28 ≤ P ≤ 516.33 N

[Torsten]

In my opinion, there is a unique value for the magnitude of the force that keeps the block in equilibrium. Why do you think that there is range for P ?

[Doug Leaffer]

There is a range for P that satisfies the equilibrium conditions. Assume, for example, that P = 0. The solution becomes simpler, and the friction force (up the ramp) is clearly > than the gravitation force (down the ramp) = equilibrium. This remains true for a range of positive P values (implies pulling), up to Pmax = 516.33 N (recall that static friction is a force that 'grows" to it's max), beyond which the block moves up the ramp. And for P <0 values (implies pushing) down to -35.28 N, at which point friction (pointing up the ramp) is overcome by the pushing force and the block moves down the ramp. The block cannot be in equilibrium at P = +572.6 N, it would move up the ramp if P > 516.33 N

[Torsten]

The block cannot be in equilibrium at P = +572.6 N, it would move up the ramp if P > 516.33 N

Yes, you made a mistake when specifying Py in your code. I corrected that (see below).

Still I cannot follow your argumentation concerning the range for P, but I'm not a physicist.

[Doug Leaffer]

Thanks for correcting the Py mistake ! Appreciate that @Torsten

Answer 3

[Paul]

Here is an algebraic solution; I suspect one could do the whole thing symbolically and then sub the numbers at the end.

Note that my expression for Fnetx is different because, to my understanding, the relative motion resisted by the static friction depends on the sign of Px - Fx, not Px (nor P!) alone. That is, we need Px > Fx for the block to want to slide up the hill and Fx > Px for the block to want to slide down the hill. Comes up with the same solutions for the parameters for this problem.

Disclaimer: I haven't done statics in more time than I care to admit.

syms P real
m = 100; % mass, kg
angle_ramp = 15; % degrees
angle_bar = 20; % degrees
g = 9.81; % gravity (m/s^2)
coeff = 0.3; % coefficient of static friction 
W = m*g;
Px = P*cosd(angle_bar); Py = P*sind(angle_bar);
N = W*cosd(angle_ramp) - Py;
Fx =  W*sind(angle_ramp);
Fs = coeff*N;
Fnetx = Px - Fx - Fs*sign(Px-Fx);
figure
fplot(Fnetx,[-100,600]),grid

●

assume(Px-Fx > 0) % motion tends up the hill
Ppos = vpa(solve(Fnetx,P))

Ppos = 

516.33324748670646195665428956607

assume(Px-Fx < 0) % motion tends down the hill
Pneg = vpa(solve(Fnetx,P))

Pneg = 

[Doug Leaffer]

Good work @Paul