# Ball- Tipping Plank

Hello, I am new to the forum and am in the process of studying for a test that I have tomorrow. The attachment that I have included is a screenshot of a quiz we have taken with the answers included already, but I am not able to figure out how the problem is done.
Any help would be appreciated since I have been trying to solve it for over an hour now.

#### Attachments

• 51.4 KB Views: 388

Related Introductory Physics Homework Help News on Phys.org
PhanthomJay
Homework Helper
Gold Member
What will be the reaction (normal) force at A when the plank just starts to tip? Use the equilibrium equations, please post your attempt and relevant equations for further assistance (per Forum rules), and Welcome to Physics Forums! What I did was:
Torque:

Relevant equation= at equilibrium:
sum of torque= 0

Attempt:
0-Mg/3(x)+ Mg(.4)(pivotb) -Mg(.5)(center of gravity)= 0.....when I solve for x I get .3, but the answer is .1

PhanthomJay
Homework Helper
Gold Member
What I did was:
Torque:

Relevant equation= at equilibrium:
sum of torque= 0
Don't forget sum of Fy =0
Attempt:
0-Mg/3(x)+ Mg(.4)(pivotb) -Mg(.5)(center of gravity)= 0.....when I solve for x I get .3, but the answer is .1
What is the reaction force at B? It is not Mg.

That's what I seem to be having an issue with. Any idea? I'm assuming it's just Force * R(distance). The force in this case I assumed was Mg. What would it be?

PhanthomJay
Homework Helper
Gold Member
Try summing forces in the vertical y direction. You know the given downward loads, and once you know what the reaction at A is, you can find the reaction at B. So, again, what must be the reaction at A when the plank is just starting to tip over and rotate about B? Hint: When it starts to tip, it loses contact with the support at A.

Wouldn't the reaction force just be 0 since there is no contact? How do I go about finding B?

After thinking about it for a while, I figured it out.

Solution question 6:

When the plank starts to pivot, it will pivot about point B. A can be neglected since it will not be in contact. Treating B as the center and the ball as the weight to the left and the center of gravity as the weight to the right, I got the following equation:

The ball will be -.4+x away from the ball and the center of gravity is at .5-
.4=.1
-Mg/3 (-.4+x) -Mg(.1)=0

Solving for x gives .1L.

Question 7:

For this part I used the original lengths given. This is the equation obtained:

-Mg/3(.1) + (x)Mg(.4) - Mg(0.5)=0

When solving for the x, 4/3 is the correct answer which corresponds to 4Mg/3.

Thank you for all of the help that lead me to the answer.

Last edited:
PhanthomJay
Homework Helper
Gold Member
Wouldn't the reaction force just be 0 since there is no contact? How do I go about finding B?
That's correct. So now just sum forces in the vertical y direction. You have the weight of the plank, mg, and the weight of the ball, mg/3, acting down. The reaction at A is 0, so what must be the reaction at B, acting up?? Then solve for x by summing torques = 0 about any convenient point (it is simplest to sum torques about B, now that you know that the reaction at A is 0).

PhanthomJay
Homework Helper
Gold Member
After thinking about it for a while, I figured it out.

Solution question 6:

When the plank starts to pivot, it will pivot about point B. A can be neglected since it will not be in contact. Treating B as the center and the ball as the weight to the left and the center of gravity as the weight to the right, I got the following equation:

The ball will be .4-x away from the ball and the center of gravity is at .5-
.4=.1
-Mg/3 (.4-x) -Mg(.1)=0

Solving for x gives .1L.

Question 7:

For this part I used the original lengths given. This is the equation obtained:

-Mg/3(.1) + (x)Mg(.4) - Mg(0.5)=0