Two moments around a point question

[IDK10]

Homework Statement

We are given a diagram to which we need to make measurements of. There is a line Y which is 9.5cm long. At one end, point X, is a force of 6,500N going perpendicular to the line; at the other end is point J, which the line continues for another 4.5cm to point P which is the pivot point.

At point J, there is a line going 60° towards point P, this is another force line, it is going on the same general direction as the other force line. All lines, except force lines are at a 1:10 scale. Using the measurements, we need to find the unknown force of the second force line.

Homework Equations

I know the sum of counterclockwise moments equal the sum of clockwise moments. Also, if it is at an angle, you use sin(thetha)x the length of the line to the pivot which is perpendicular to the line of action of the force.

The Attempt at a Solution

I tried find the moment of the kniwn force, 6,500 x (9.5x10)/100= 910Nm. Then I used 910 + w₂x₂ = 0

w₂0.21 [From 2.1x(100/10). Also, I got 2.1 from measuring the length of the line from P which meets up with the line perpendicular to that of the force]₂ = -910
-910/0.21=4,333.333... (I know its meant to be negative, but I didn't think a negative force would work)

 

[IDK10]

Can someone also tell me why this got moved to introductory phsyics? We do this at Year 12.

 

[kuruman]

Second post gets answered first. The level of this problem has been correctly evaluated as introductory physics. This is "introductory" at College or University level not primary education.

Now for the first post. A clarification would help. The question as you posted it states

Using the measurements, we need to find the unknown force of the second force line.

You do not specify what the system is doing. I see two possibilities:
1. The system is at equilibrium. In this case a diagram is needed because the way you describe it, both torques are in the same direction and the system cannot be in equilibrium.
2. The system is accelerating. In this case the acceleration is needed to find the unknown force.

Can you post the problem the way it is originally phrased?

 

[IDK10]

Second post gets answered first. The level of this problem has been correctly evaluated as introductory physics. This is "introductory" at College or University level not primary education.

Now for the first post. A clarification would help. The question as you posted it states

You do not specify what the system is doing. I see two possibilities:
1. The system is at equilibrium. In this case a diagram is needed because the way you describe it, both torques are in the same direction and the system cannot be in equilibrium.
2. The system is accelerating. In this case the acceleration is needed to find the unknown force.

Can you post the problem the way it is originally phrased?

I wasn't stating the question, I was describing the layout of the arrow, I'll see if I can scan it. We are just given one of two known forces. And its not in equilibrium. Basically, there is a 12cm (1.4m with 1:10 scale) line connected to a pivot. At the end, a 6,500N force is pulled on it. An unknown force also acts 4.5cm (0.45m with 1:10 scale) from the pivot acting 60° towards the pivot.

Both moment are clockwise, none are anticlockwise.

 

[kuruman]

Is this what you have and are looking for the unknown force F?

 

[IDK10]

Is this what you have and are looking for the unknown force F?

View attachment 112369

Yes, saves me time, mine is rotated slightly and has numbers and lines drawn n it, which I did for measurements.

 

[IDK10]

Is this what you have and are looking for the unknown force F?

View attachment 112369

And, X-J = 9.5cm
J-P = 4.5cm
P is the pivot point.

Also, the Force at F must provide the 6,500N

 

[kuruman]

OK. Now can you post the task that you are expected to perform exactly as it was given to you?

 

[IDK10]

OK. Now can you post the task that you are expected to perform exactly as it was given to you?

How do I upload a scanned image?

 

[kuruman]

Use the "UPLOAD" button, near the bottom right of your screen.

 

[IDK10]

Use the "UPLOAD" button, near the bottom right of your screen.
View attachment 112371

Taken from the word document.

 

[kuruman]

I see what's going on now. It looks like you got the wrong impression from the figure. There are two forces that generate torques on the lower jaw about the pivot at P. (1) the unknown muscle force that is up and to the right as shown in the figure; (2) the tooth force of 6500 N that is down opposite to the arrow that is shown in the figure. In short, the muscle exerts a clockwise torque on the lower jaw while the upper jaw exerts an anticlockwise torque which keeps the lower jaw in static equilibrium and clamped shut. So, to solve this problem, reverse the direction of the 6500 N force and then balance the torques. That should work.

 

[IDK10]

I see what's going on now. It looks like you got the wrong impression from the figure. There are two forces that generate torques on the lower jaw about the pivot at P. (1) the unknown muscle force that is up and to the right as shown in the figure; (2) the tooth force of 6500 N that is down opposite to the arrow that is shown in the figure. In short, the muscle exerts a clockwise torque on the lower jaw while the upper jaw exerts an anticlockwise torque which keeps the lower jaw in static equilibrium and clamped shut. So, to solve this problem, reverse the direction of the 6500 N force and then balance the torques. That should work.

That's what I basically did in the attempt at a solution section.

 

[kuruman]

This equation

910 + w2x2 = 0

should be 910 - w2x2 = 0 because the torques have opposite signs. The quantity w2 is positive because it represents the magnitude of the force. Its direction is taken into account and built in the equation when you figure out if the torque is clockwise or not. Is this what bothered you?

 

[IDK10]

This equation

should be 910 - w2x2 = 0 because the torques have opposite signs. The quantity w2 is positive because it represents the magnitude of the force. Its direction is taken into account and built in the equation when you figure out if the torque is clockwise or not. Is this what bothered you?

Yeah, I mean, they are on the same side.

 

[kuruman]

Yeah, I mean, they are on the same side.

The forces acting on the lower jaw are not on the same side. If you were to draw a free body diagram of the lower jaw showing all the forces that generate torques about the pivot it should look like this

If you don't believe me, stick your finger between your front teeth and bite down (not too hard). You will feel your upper teeth exert a down force on the upper part of your finger. If you remove your finger, and bite, that same down force will be exerted on your lower jaw. That down force is the 6500 N in this example. The picture of the T. Rex skull that you posted is misleading in that regard.