# Forces on a painting

1. Aug 14, 2008

### shanie

1. The problem statement, all variables and given/known data
This question comes from a lab that we did a while ago: Hang a painting on a nail by a thread. The weight of the painting was 0.4 kg.
a) Calculate the force in the nail that holds up the painting.
b) Calculate the forces that work on the thread holding the painting in place.

2. Relevant equations
Newton's second law, F=mg
And the normal force that is as large but working in the opposite direction, since the painting is in equilibrium.

3. The attempt at a solution
a) When attempting to solve this question I observed the phrasing "in the nail", does this mean that I have to weigh the nail and then calculate the gravitational force, and then the normal force from the wall that holds the nail in equilibrium? Or does this mean that I have to use the weight of the painting including the thread and then make this the force in the nail?
b) Now what I did was draw the following picture

http://img182.imageshack.us/img182/382/tavlan6zw8.png [Broken]

I made the assumption that the gravitational force works from the centre of the painting, and that the normal force works on the painting from the thread in two points in the opposite direction (i.e., the points where the thread holds the painting)..

I'm having trouble coming to terms with what forces work on which objects.. I could really use some help, thanks!

Last edited by a moderator: May 3, 2017
2. Aug 14, 2008

### stewartcs

For some reason I can't see your picture. However...

In the first part, I imagine they are asking what is the force acting on the nail as a result of the painting hanging from it by the wires (thread). Start with drawing a FBD. Based on the problem description, it should be safe to assume the wires are at the same angle which will simplify the solution some.

In the second part the force in the wires (thread) is just the tension due to the painting hanging. Your FBD will show you what forces are acting on the system. HINT: The since the angles are the same for the wires the tensions will be the same.

CS

Last edited by a moderator: May 3, 2017
3. Aug 14, 2008

### shanie

a) What's an FBD? So I should just disregard the phrasing and assume they mean that I should find the force ON the nail? So then I just set F=mg=0.6*9.82=5.9N? And then do the same for the normal force on the nail which is of the same magnitude but in the opposite direction?

b) So each wire will have half of the normal force upwards, ~2.9N in each then?

4. Aug 14, 2008

### stewartcs

Still can't view the picture, but I know why I can't now, my company blocks that website for some reason.

Anyway, I would use the initial assumptions we discussed previously, they are typical for these types of problems.

FBD stands for Free Body Diagram. It will help you see all of the forces acting on the system so you can ensure that everything is accounted for in your equilibrium equation(s).

I just noticed the problem didn't state what angles the wires made. Was this not indicated in the lab? You'll have to know that in order to find the tension in each wire since it is a function of those angles.

After you draw your FBD, apply Newton's Second Law to come up with equations that relate the forces.
HINT: You will need to break the system down into x and y components.

Finally, the net force acting on the nail will be the sum of the vertical componets of the tension in the wires supporting the picture.

CS

5. Aug 14, 2008

### shanie

I realize now you meant free body diagram, so what I figured out was the following:
a) The forces on the nail is the gravitational force from the wire and the painting, pointing downwards. The nail works on the wire by a normal force that is as large as the gravitational force 5.9N but in the opposite direction. These forces cancel out as the nail is in equilibrium.
b) The forces on the wires come from the painting's gravitational force of 5.9N along with the normal force from the nail which is equivalent in the upward direction, keeping the wires in place. The normal and gravitational force is equally divided between the wires, giving them half of it each. Hence, about 2.9N works in each wire.

I would really appreciate if you could tell me if this is correct?

6. Aug 14, 2008

### stewartcs

The equilibrium equation you should have come up with based on your FBD should be:

$$T_1 + T_2 - F_g = 0$$

This takes the painting as the system and looks at the forces acting on it. Namely, the tension in each wire (T1 and T2) and gravity (Fg which equals mg).

Since the painting is not moving, the RHS of the equation is 0.

Now break down the forces into components (assume the angle is the same):

x: $$-T_1 \cos{\theta} + T_2 \cos{\theta} - 0 = 0$$

which reduces to T1 = T2

y: $$T_1 \sin{\theta} + T_2 \sin{\theta} - mg = 0$$

which reduces to:

$$T_2 = \frac{mg}{2\sin{\theta}}$$

Note that the tension in the wires varies with the angle. If you assume the angle is 45 degrees, then you'll get 2.77 N in T1 and T2 (since T1 = T2 at the same angle), which is close to what you came up with. Vary the angle using these equations and see how your tension changes.

Also, you used 0.4 kg for the mass in the original problem and 0.6 kg when you determined mg in your second post. The 2.77 N answer assumes the mass of the painting is 0.4 kg as first posted.

The vertical force on the nail is simply equal to Fg which equals mg, or 3.92 N. Again, assuming 0.4 kg was the correct mass (i.e. 0.4 x 9.81 = 3.92). That is to say that the sum of the vertical components of the tension in the wires is 3.92 N. You can check this by taking T1 and T2 and find the sum of the vertical components of tension (i.e. $$T_{y,total} = T_1 \sin{\theta} + T_2 \sin{\theta}$$ ). Which works out to equaling mg.

Hope this helps.

CS

7. Aug 14, 2008

### shanie

Thank you, that really helped! I was just wondering, does the nail exert an upward normal force on the wire (seeing as another force must help keep the wire and painting in equilibrium)? And if so, then is it, in this case, equal to the gravitational force since it must be perpendicular to the surface?

8. Aug 14, 2008

### stewartcs

Yes, the nail will have an equal but opposite reaction force that is equal to mg in this case. The force will be perpendicular to and at the point the wires connect to the nail.

CS