Math Is Fun Forum

jimi70

Member

Registered: 2012-04-04

Posts: 22

Block on horizontal plane

5 N
     \         |
         \60 deg
              \|________
               |              |
  4 N ----- |________|

A rectangular block B of weight 12 N lies in limiting equilibrium on a horizontal surface. A horizontal force of 4 N and a coplanar force of 5 N inclined at 60 degrees to the vertical act on B.

(i) Find the coefficient of friction between B and the surface.

B is now cut horizontally into two smaller blocks. The upper block has weight 9 N and the lower block has weight of 3 N. The 5 N force now acts on the upper block and the 4 N force now acts on the lower block (see diagram below)

5 N
     \         |
         \60 deg
              \|________
               |_9 N____|
  4 N ----- |_3 N____|

The coefficient of friction between the two blocks is u.

(ii) Given that the upper block is in limiting equilibrium, find u.

(iii) Given instead that u = 0.1, find the accelerations of the two blocks.

my answers:

(i) (4 + 5 sin 60) / (12 - 5 cos 60) = 0.877
(ii) (5 sin 60) / (9 - 5 cos 60) = 0.666
(iii) for the upper block: 5 sin 60 - 0.1(9 - 5 cos 60) = 9a/9.8
so a = 4.01 m/s/s

I don't know how to calculate the acceleration for the lower block as I don't know how the friction between the two blocks affects the tractive force of the lower block. Can somebody help me?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,140

Re: Block on horizontal plane

hi jimi70

Diagram for part (i) below.

EDITED VERSION.

I agree with your answer to (i).

Part (ii) coming up in the next post (if I can work it out   )

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,140

Re: Block on horizontal plane

Now for the second part.

EDITED TO TAKE ACCOUNT OF THE CORRECT FORCE DIRECTIONS.

It's the reactions that make this trickier for me.

You need to consider each block separately, making sure you only take account of forces that act on that block.

I've tried to colour code the forces, red for the top block, and pinky/purple for the bottom block.

I've called the top friction, Ft and the bottom friction Fb

Rt is the force that the bottom block exerts upwards on the top block.

It is also the force that the top block exerts downwards on the bottom block.  Use this instead of the 9N weight because that acts at the centre of the top block; not directly on the bottom block. (again the 5N component and the 9N weight act together one to add to the downwards force and the other opposiing this on the bottom block, and that's what I'm calling Rt.)

Rb is the reaction of the ground upwards on the bottom block.

In my diagram, I've shown the reactions offset to the right of the weights, so that these vertical forces can be appreciated more easily.  In reality the reactions act through the centre of gravity too.

So in limiting equilibrium for the top block

............................................................................................................

And if u = 0.1

m = 9/g

Use Rt unchanged from the above to get this acceleration.

.............................................................................................................

For the bottom block:

m = 3/g

That should allow you to find the acceleration for the bottom block.

You'll have to assume that 'u' has the value found at the beginning of this problem.

Post back your answers and I'll check them.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

jimi70

Member

Registered: 2012-04-04

Posts: 22

Re: Block on horizontal plane

bob

sorry, it's my diagram. The 5 N force is supposed to be acting upwards at an angle.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,140

Re: Block on horizontal plane

Arhh! OK.

But all you have to do is make all my 5s into -5s.  The rest still stands

Your answer (i) still needs to be adjusted as you have both 4 and 5sin60 acting the same way.

Wait a minute though.

4 - 5sin60 = -0.33.... so the frictional forces act the other way.

As my Fs are just algebraic symbols if you leave my equations unchanged they'll come out negative.  ie they act the other way.

Same goes for the accelerations.

When I get a longer moment, I'll go back and edit posts 2 and 3 to reflect this change.

Oh!  I've just noticed something else.  Which way is the 4 acting?  I have assumed towards the right.

Before I change anything we'd better agree we are both doing the same question.

So the 5 acts up.

Which way the 4 ?

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

jimi70

Member

Registered: 2012-04-04

Posts: 22

Re: Block on horizontal plane

bob

sorry again. The 4 N force acts in the left direction of the block.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,140

Re: Block on horizontal plane

hi jimi70

Not to worry.  I should have considered that at the start and not made assumptions.

I have edited both posts and diagrams to show these forces correctly.

The equations hardly change, in fact, and you'll be happy to know I now agree with your first answers. (Should have realised I'd got the directions wrong shouldn't I.  Whoops, sorry.)

I've replaced 9 and 3 in the accelerations with 9/g and 3/g.  Double whoops!

Hopefully it is now possible for you to complete the question.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

jimi70

Member

Registered: 2012-04-04

Posts: 22

Re: Block on horizontal plane

bob

Just a couple of questions about your second diagram:
Is that right that the friction acting against the top block, Ft, acts in the opposite direction on the lower block?
Shouldn't Fb be acting in the opposite direction? (It's probably my diagram. I'll try and download an image next time).

btw, thanks for all your help.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,140

Re: Block on horizontal plane

Hi jimi70

As the top block (let’s call it T) drags across the bottom block (B) , the friction with B is trying to stop it.  So the friction that B exerts on T must be against the motion;  ie. to the right.
At the same time T is trying to speed up B’s movement by dragging it to the left.  So an equal force is exerted by T on B.  It will always be like this; whatever T does to B, B does an equal and opposite amount to T.  Notice I have colour coded the T forces in red and the B forces in pinky/purple to help identify which forces to use for each equation and so you won’t get them muddled.

There is one extra bonus in using algebraic symbols for the forces.  If we put a force the wrong way (I’m not talking about the given forces but rather the ones we introduce as letters) the algebra will simply allow  it to be negative and the final results will be unchanged.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob