Determine forces acting on a plank at rest

[zeralda21]

Homework Statement

A plank with length L is held in a place by two supports, one plank at one end A and the other a distance 5L/12 from the same end B, and is otherwise only affected by gravity. The plank has mass m. How can the vertical force on the plank be determined?

Homework Equations

I think this: Moment=Force*Distance. Here distance is the closest from a given moment point to a force's "point of application.

The Attempt at a Solution

I have noticed that the supports are placed at opposite sides of each other and creates a torque that are trying to rotate the plank to different directions. So in order for the plank to be at rest, the sum of the total torque must be zero.

There are two forces, a downward vertical force at A, and a upward force at B. For the plank to be in equilibrium they must be equal, or Fnet_{y}=0. Correct?

I'm not sure which moment point to choose but let it be B. Then M=A(5L/12). But it doesn't help much. And where do mg come in? i know there's a gravitational force mg but i cannot fit it in any equation.

 

[cepheid]

You need to revise your free body diagram for the plank. There are THREE vertical forces that act on it. One is gravity, mg, downward, which acts *at the centre of mass.* The other two are the two upward reaction forces (contact forces) that act at point A and at point B, to support the weight of the plank. These are both unknown

Force equilbrium alone will allow you to reduce this to one unknown (because you can express F_A in terms of F_B or vice versa). All the force equilibrium tells you is that the *sum* of the upward forces must balance the downward one. It does not allow you to solve for the two upward ones individually.

Balance of torques is required to solve for the remaining unknown. It does not matter around what point you compute the torques, as long as you are consistent. Personally, I would choose the centre of mass, so that you only have to compute two torques, not three.* (Gravity produces zero torque around the centre of mass, because its line of action passes through that point).

*EDIT: I guess that would be true of either point A or point B as well. Duh.

 

[PhanthomJay]

Homework Statement

A plank with length L is held in a place by two supports, one plank at one end A and the other a distance 5L/12 from the same end B, and is otherwise only affected by gravity. The plank has mass m.

I think you mean
"A plank with length L is held in a place by two supports, one support at end A and the other support at point B. Point B is a distance 5L/12 from A. The plank is otherwise only affected by gravity. The plank has mass m."

How can the vertical force on the plank be determined?

I think you mean "How can the vertical forces of the supports on the plank be determined?"

Homework Equations

I think this: Moment=Force*Distance. Here distance is the closest from a given moment point to a force's "point of application.

The Attempt at a Solution

I have noticed that the supports are placed at opposite sides of each other and creates a torque that are trying to rotate the plank to different directions. So in order for the plank to be at rest, the sum of the total torque must be zero.

the total torque about any point must be zero, and that includes the torque from the beam's weight acting at its cg.

There are two forces, a downward vertical force at A, and a upward force at B. For the plank to be in equilibrium they must be equal, or Fnet_{y}=0. Correct?

A is up and B is down, yes, and don't forget the weight acting down.

I'm not sure which moment point to choose but let it be B. Then M=A(5L/12). But it doesn't help much. And where do mg come in? i know there's a gravitational force mg but i cannot fit it in any equation.

Choosing B as your point is OK, then sum torques of both the force at A and the weight about that point = 0 to solve for A.

 

[cepheid]

Hey, sorry I wasn't paying close attention and missed the fact that support B was less than half-way across the beam. The basic idea still applies.

 

[zeralda21]

Hey, sorry I wasn't paying close attention and missed the fact that support B was less than half-way across the beam. The basic idea still applies.

Oh, excellent posts. I was actually very confused that there was an upward force at A but I carried on anyway and chose the centre of mass as my moment point. Anyway, the result only had wrong sign; My calculations gave -mg/5 while the correct was mg/5. But that is, of course, because I chose wrong sign for the force at A. Thanks a lot.

Ps. I see that I haven't mentioned that the supports are not on the same side, hence the sign. I take full blame of this.

 

[cepheid]

Oh, excellent posts. I was actually very confused that there was an upward force at A but I carried on anyway and chose the centre of mass as my moment point. Anyway, the result only had wrong sign; My calculations gave -mg/5 while the correct was mg/5. But that is, of course, because I chose wrong sign for the force at A. Thanks a lot.

You're welcome. Yeah, that's the nice thing about these statics problems. It doesn't matter what direction you initially assume for a force. If your assumption is "wrong" you'll get a negative answer, which is still correct. A "negative upward force" is a downward force.

Ps. I see that I haven't mentioned that the supports are not on the same side, hence the sign. I take full blame of this.

You DID mention it. It's mentioned explicitly in the problem statement. After all: (5/12) < (1/2). I should have noticed this and I am to blame.