# Homework Help: 2D Forces Equilibrium Question

1. Oct 10, 2008

### Melawrghk

1. The problem statement, all variables and given/known data
The mass of block A is 42kg, and the mass of block of B is 50kg. The surfaces are smooth. If the blocks are in equilibrium, what is the force F?

2. Relevant equations
$$\sum$$Fx=0
$$\sum$$Fy=0

3. The attempt at a solution
I added the vectors do this diagram. The normal force to both blocks, the weight of each box. I treated the two boxes as one whole so I wouldn't have to worry about their contact area.

I got this for my equations:
$$\sum$$Fx= NA*cos(70) + F - NB=0
$$\sum$$Fy= NA*sin(70) - 42*9.81-50*9.81=0

I can solve for Na, but I can't ge to F. Help?

Last edited: Oct 10, 2008
2. Oct 10, 2008

### Rake-MC

Interesting, NA this is the normal reaction force of box A? Also, both of your equations are sums of forces in the x direction..

3. Oct 10, 2008

### Melawrghk

Oops, sorry, that's a typo. The first one is x, second one is y. Na is just the normal force exerted by the surface on A in response to its weight.

4. Oct 10, 2008

### Rake-MC

Are you sure your formulae are right?

First re-write NA as $$M_agcos(20)$$

Then you can solve your forces in y.
Try this first.

5. Oct 10, 2008

### Melawrghk

Well actually I could solve my forces in y already, so that's not such a problem. My problem rests with not knowing anything about Nb.

I'm fairly sure about my equations.

6. Oct 10, 2008

### Rake-MC

Perhaps before being so sure about your equations, you should try what I said in my last post.

7. Oct 10, 2008

### Melawrghk

Well, I don't think I can say that Na=Ma*g*cos(20) because it's not just A acting at that point. Object B also exerts a force on that point in one or another. Plus there is a force being applied to the whole system.

Using your logic, no, my equations aren't right. But this is how I was taught to treat normal forces in 2D force systems...?

8. Oct 10, 2008

### Rake-MC

But according to this, my logic is the one that you appear to be using.

All this aside, the point I am making is that you have not taken into account, the normal reaction force of mass b.

9. Oct 10, 2008

### Melawrghk

Reading into words aside, that is obviously not the approach I'm using as I just explained in the post above. I also stated that I'm treating two objects as one, meaning normal force exerted on that contact side of A is the normal force exerted on the whole system from that point. I did take normal force acting on B-side and that is NB. The force acts perpendicular to the surface (ie in the x-direction only).

Sorry if any of this sounded harsh/rude, it honestly wasn't intended that way. I'm just trying to get this question and not be critiqued on my explanation/description mis-skills.

10. Oct 10, 2008

### Rake-MC

Yeah my apologies too, when you said 'Na is just the normal force exerted by the surface on A in response to its weight.' You actually meant "Na is just the normal force exerted by the surface on A in response to its weight + the weight of B"

Okay well we can look at this slightly differently then.
You need to calculate the amount of force B exerts on A. B has mg down. It is at 45 degree angle with A.
Therefore the normal force between B and A is mgsin(45). Once you have that, you can calculate the vertical component of that and add it to the y force of A.

11. Oct 10, 2008

### Melawrghk

But where does that get me? I got the vertical component of that equalling 245.25N and if I add it to the Fy, Na becomes 699N. What then?

12. Oct 10, 2008

### Rake-MC

Well, it does two things for you. One, it gives you the numerical value of Na, which will come in handy for solving the equation soon.

Two, it allows you to calculate the horizontal force on B due to gravity.