# Find minimum force to move block

## 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.

#### Attachments

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Simon Bridge
Homework Helper
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.

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

Simon Bridge
Homework Helper
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
Homework Helper
Conservation of energy means: ##E_{in} = E_{out} + E_{lost}##.

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