Recent content by applesoranges

Upwards.

The angle between T1 and horizontal will have to increase to make sure mac equals T1x.

Not a all, just a brain freeze. Good question, because now I understand that T2 can't really be negative. I mean, algebraically you could make T2 in ##T_{2}=\frac{T_{1}sin\theta-mg}{sin\theta }## less than zero. But that situation can't physically exist, right? When the string is cut and T2=0...

Given: T1=22.0N m=0.500 kg ##\theta = 30.0^{\circ}## Find: T2 = ? What I have so far: ##\sum_{}^{}F_{x}^{}: T_{1}cos\theta +T_{2}cos\theta =ma_{c}## ##\sum_{}^{}F_{y}^{}: T_{1}sin\theta =T_{2}sin\theta + mg## ##T_{2}=\frac{T_{1}sin\theta-mg}{sin\theta }## The result for T2 is a positive number...

What throws me off in the summation of forces in y-direction is T2y vector being opposite in sign to T1y. I get a negative number for T2y... I'm trying to comprehend what to make of it. I'll try to work on it more when I come back after my night shift.

What throws me off in the summation of forces in y-direction is T2y vector being opposite in sign to T1y. I get a negative number for T2y... I'm trying to comprehend what to make of it. I'll try to work on it more when I come back after my night shift.

Fair enough! I guess I'll have to model a situation where v is the way that T2 is 0 and see how it goes...

I have already solved this problem, just would like to double check something with you conceptually. I've got a negative result for the tension in the lower cord. Intuitively I think it is right, because the lower cord does not support the ball in its opposing the force of gravity. It actually...