Find minimum force to move block

[the_sense]

Homework Statement

"Figure 1 shows two wedge blocks A and B, weighing 1000 N and 50 N, respectively, in contact. The coefficient of friction between all contacting surfaces is 0.1 . What would be the smallest value of P required to push the block A upwards? What would be its value if there were no friction?"

Ans, with friction: 887.3 N (given)
Ans, without friction: 577.4 N (given)
The gray rectangle in the picture is part of a separate problem.

Homework Equations

$ƩF = 0$

$F_f = μF_N$

The Attempt at a Solution

I have solved problems of this nature before. I have a good understanding of vector addition and statics. I solved the problem sans friction in the following manner:

$F_N = W_A /cos30$

$ƩF_x = 0 ∴ F_N \cdot cos60 = P$

$P = W_A \cdot cos60 /cos30 = 577.4 N$

With friction, however, I run into a problem: The side walls in contact with A exert a frictional force downwards. This of course depends on the normal force exerted by those walls, which partly balance the horizontal components of the normal and frictional forces exerted by B on A. But as far as I can tell, the vertical component of the normal force is equal to the sum of the weight and these frictional forces. So I encounter what appears to be a catch-22. I have tried various methods employing systems of equations but I get nowhere.

 

[Simon Bridge]

I would try the work-energy relation and conservation of energy.
Otherwise you'd have to draw the free body diagrams.

The wrinkle I see is that block A also gets pressed against the side of the whatsits that constrains it to the vertical - in proportion to the applied force. (The applied force ends up creating a couple with the reaction forces in the constraint - I suspect that this is "overthinking" the problem at the level it is presented.)

You should get a number of equations where it appears that things have a circular dependence - and you can cancel terms.

 

[the_sense]

Ok, I will try that. Thanks for the help.

 

[Simon Bridge]

At least you have nice angles:

sin(30)=1/2
cos(30)=(√3)/2
tan(30)=1/√3

... you can replace the trig functions with the fractions (keep the surd until the end).
Most of them seem to come out to 1/2.
In all your triangles for this problem, find the 30deg angle and do all your trig with that as theta
... you will get less confused.

 

[Simon Bridge]

Conservation of energy means: $E_{in} = E_{out} + E_{lost}$.

Applied force P over distance x, energy input: E_in = Px.

This energy goes to gravitational PE, $U$, in A, and three lots of friction:
E₁: between B and the ground
E₂: between the blocks
E₃: between A and the restraining channel.

So: $E_{in}=U+E_1+E_2+E_3$

$U=w_A\Delta y:w_A=m_Ag=1000\text{N}$
B moving x means A moves up $\Delta y=x\tan(30^\circ)=x/\sqrt{3}$
$\Rightarrow U=\frac{1}{\sqrt{3}}w_Ax$

Without friction: $P=\frac{1}{\sqrt{3}}w_A$

... which is where you are up to ;)


As you have observed, the friction is the tricky bit.

The ith energy lost to friction is $E_i=f_id$ where d is the distance moved against friction and $f_i=\mu N_i$ ... so you are finding expressions for the "normal force", $N_i$ ... which will mean drawing free-body diagrams. Assuming flush surfaces will probably avoid having to think about couples.

i.e. For $i=1$ (the first friction): $d=x$, and $N_1 = w_B+\text{<contributions from }w_A\text{>}$

Try to express each normal force in terms of P and x and constants, where applicable.
You'll end up with something like $Px=\big(p(P)+q\big)x$ where you can eliminate x and solve for P.