Box Slipping and Tipping: Finding the Critical Force P

[zeralda21]

Homework Statement

We have a homogenous box on a surface with static friction μ=0.5. We apply a force on the top of the box at an angle 30 degrees to the horizontal (see picture) and we want to know for what values of P does the box slip and tip. I have drawn a free-body diagram here of the situation:

http://i.imgur.com/jRh1zqa.png

The Attempt at a Solution

I have found that for slipping, P=0.448mg but I need help to determine P for the box to tip. If we take the moment(torque) around point C we get;

(Pcosθ)d+(Psinθ)2d+(N_B)2d-mgd=0 and hence P = (mg-2N_B)/(cosθ+2sinθ). How can I get rid of N_B in this case?

 

[darkxponent]

Hint: you missed a torque in the torque equation. Just have a look at the FBD again.

 

[zeralda21]

Hint: you missed a torque in the torque equation. Just have a look at the FBD again.

Do you mean the friction force F_B? Because its line of action is parallel and pass through C and therefore no moment.

 

[darkxponent]

oh sorry i didn't see the mgd in equation. So you have taken all the Forces that give torque but still the equation is incorrect. Just see whether the perpendicular distances you have taken are correct?

Correcting it still won't give you the answer. Now focus on N_B. Hows is tipping going to effect it.

 

[zeralda21]

oh sorry i didn't see the mgd in equation. So you have taken all the Forces that give torque but still the equation is incorrect. Just see whether the perpendicular distances you have taken are correct?

Correcting it still won't give you the answer. Now focus on N_B. Hows is tipping going to effect it.

N_B=0 if it is tipping right? So P = (mg)/(cosθ+2sinθ)) which is correct.

 

[darkxponent]

No N_B is not zero when it is tipping. Normal becomes zero only when object leaves the surface. But the answer you got is still correct!. Now try too think what happens to N_B at tipping?

 

[haruspex]

No N_B is not zero when it is tipping. Normal becomes zero only when object leaves the surface. But the answer you got is still correct!. Now try too think what happens to N_B at tipping?

I think Zeralda is defining N_B as the normal force acting at B (and presumably has N_C for a normal force at C, but that has no moment about C). With that definition, N_B does go to zero.

 

[darkxponent]

I think Zeralda is defining N_B as the normal force acting at B (and presumably has N_C for a normal force at C, but that has no moment about C). With that definition, N_B does go to zero.

I don't know how to define normal at B and C separately. I thought N_B is the net normal at Base. And that what Zeralda has shown in the FBD( Normal is in the middle of the objects Base). I was trying to say that N_B shifts at point C at tipping and hence its moment is zero.

I also said Zeralda has done some mistake in taking the perpendicular distances. I was talking about N_B, it should be (N_B*d) not (N_B*2d). As we always take normal at the middle of the surface.

Correct me if i am wrong somewhere. I am just a student.

 

[haruspex]

I don't know how to define normal at B and C separately. I thought N_B is the net normal at Base. And that what Zeralda has shown in the FBD( Normal is in the middle of the objects Base). I was trying to say that N_B shifts at point C at tipping and hence its moment is zero.

I also said Zeralda has done some mistake in taking the perpendicular distances. I was talking about N_B, it should be (N_B*d) not (N_B*2d). As we always take normal at the middle of the surface.

Correct me if i am wrong somewhere. I am just a student.

I based that on the diagram, which shows separate frictional forces at B and C. It also explains the use of 2d as the multiplier.
In practice, the normal force is distributed over the base. There are two ways to represent this (for the purposes of this question). We can do it as one force which acts in different positions according to the force P, or as two forces, a normal at each of B and C, that vary in relative magnitude. Either way, we agree that on point of tipping the normal force will be entirely at C.