- #1
Resmo112
- 45
- 0
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?
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?