Find Force Applied to Tooth with Tension of 2.0 N

[Resmo112]

The drawing below shows an elastic cord attached to two back teeth and stretched across a front tooth. The purpose of this arrangement is to apply a force to the front tooth. (The figure has been simplified by running the cord straight from the front tooth to the back teeth.) If the tension in the cord is 2.0 N, what are the magnitude and direction of the force applied to the front tooth?
N toward the back of the mouth








Closest I can figure this should be the T=2(sin/cos33) depending on whether you're trying to find the vertical or horizontal component of this problem. I believe I'm trying to find the x value so I have

2=T(cos33)(2) so I solve that algebraically and I get this

2/2cos(33) = t I put that in and it comes out to 1.2 to two sig figs. but apparently that's wrong, the only other thing I can figure is the 2's cancel and I'm left with 1/cos33 = T which is also 1.2. So where am I going wrong are there more forces? should I be using sin rather than cos? or am I not even solving the right problem?

 

[tiny-tim]

Hi Resmo112!

Closest I can figure this should be the T=2(sin/cos33) depending on whether you're trying to find the vertical or horizontal component of this problem. I believe I'm trying to find the x value so I have

2=T(cos33)(2) so I solve that algebraically and I get this

2/2cos(33) = t I put that in and it comes out to 1.2 to two sig figs. but apparently that's wrong, the only other thing I can figure is the 2's cancel and I'm left with 1/cos33 = T which is also 1.2. So where am I going wrong are there more forces? should I be using sin rather than cos? or am I not even solving the right problem?

Why are you dividing by cos?

Draw a free body diagram for the forces on the teeth …

it should show only two external forces, from the tension …

the resultant force on the teeth is what you have to find.

 

[Resmo112]

Hi Resmo112!


Why are you dividing by cos?

Draw a free body diagram for the forces on the teeth …

it should show only two external forces, from the tension …

the resultant force on the teeth is what you have to find.

I thought you were supposed to, when they gave you the Newtons for tension the equation is something like T=cos*2 I solved this problem like that

A crow sits on a clothesline midway between two poles as shown. Each end of the rope makes an angle of θ = 23° below the horizontal where it connects to the pole. If the combined weight of the crow and the rope is 10.2 N, what is the tension in the rope?

and here's the solution

2 (T sin 23) = 10.2 N
T = 10.2 N / 2 sin 23
T = 13.05 N

 

[tiny-tim]

Hi Resmo112!

A crow sits on a clothesline midway between two poles as shown. Each end of the rope makes an angle of θ = 23° below the horizontal where it connects to the pole. If the combined weight of the crow and the rope is 10.2 N, what is the tension in the rope?

Yes, if they give you the weight and ask for the tension, you divide by cos (or sin),

but here they give you the tension and ask for the weight.

(if you draw the free body diagram, it should be much clearer )

 

[rwisz]

I would first clarify what the question is asking for. Is it asking for the magnitude and direction of the force applied to the front tooth, or is it asking for the force applied to the back teeth? In the drawing, it appears that the tension in the cord is pulling the front tooth towards the back of the mouth. So, if the question is asking for the force applied to the front tooth, then the magnitude and direction would be 2.0 N towards the back of the mouth. However, if the question is asking for the force applied to the back teeth, then the magnitude and direction would be 2.0 N towards the front of the mouth. It is important to clarify this in order to accurately solve the problem.

Assuming the question is asking for the force applied to the front tooth, we can use the equation F = Tcosθ, where F is the force applied, T is the tension in the cord, and θ is the angle between the cord and the horizontal. In this case, θ = 33 degrees. Plugging in the values, we get F = 2.0 N * cos33 = 1.4 N. So, the magnitude of the force applied to the front tooth is 1.4 N, and the direction is towards the back of the mouth.

It is also important to note that there may be other forces acting on the front tooth, such as the force of gravity or any other external forces. Without more information, it is difficult to accurately determine the net force on the front tooth. However, based on the given information, we can determine the force applied by the tension in the cord.