# Dynamic forces with radial acceleration

1. Sep 23, 2006

### trajan22

well this problem is killing me and ive spent a good 3 hours or so on it and still have no real results.

This is the problem
The 4.00-kg block in the figure View Figure is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are extended as shown in the diagram and the tension in the upper string is 80.0 N.
What is the tension in the lower cord?

How many revolutions per minute does the system make?

there is the picture provided

what i did

i realized that the sum of the forces in the x direction is equal to Tsin(theta) and that this is equal to the mass times the radial acceleration
I bisected the triangle in order to make it a right triangle. from here i found the angle measures using arcsin(1/1.25)... i also found the radius by using the pythagorean thrm. to be .75m.
from here i tried to solve part b but came out with a wrong answer

for the y direction i know the sum of the forces is equal to cos(theta)+(-mg)
these forces that ive described are only for the top string and not the bottom...im not really sure where to go from here or if im even on the right track

sorry to bother you guys again im just on the verge of giving up on this problem

I think the main problem is that im having troubel with finding all the forces for centripetal acceleration.
ive attached a force diagram that i made. but im missing something??

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Last edited: Sep 23, 2006
2. Sep 23, 2006

### trajan22

3. Sep 24, 2006

### big man

You know that the vertical force from the upper cord is going to be $$T_1cos \theta$$. You also know that this needs to cancel the downwards force from the wieght of the block and the tension from the lower cord.

You are right in equating the sum horizontal parts of the tension forces to the centripetal force of $$\frac {mv^2} {r}$$, but you just made a mistake in your calculations of the radius.

Like you said use pythagoras's theorem.

Use $$c^2 = a^2 + b^2$$
you know that a=1 and c=1.25 so plug that into the equation above and rearrange to solve for b.

You can then solve for v using the method you were using before (paragraph 2) and then you just need to think of how you can get the number of revolutions per minute from knowing the velocity.

4. Sep 24, 2006

### trajan22

well i still am not getting the right tension in the lower string though...i am taking Tcos(37) to be the normal force and i am subtracting the weight because it is providing slack to the lower string. where is this going wrong

5. Sep 25, 2006

### big man

So is your equation for the vertical components this:

$$T_1 cos \theta_1 = T_2 cos \theta_2 + Mg$$

where $$\theta_1=\theta_2$$

That should give you the right answer. There is one upwards force, which is $$T_1 cos \theta$$ and then there are two downwards forces, which are Mg and $$T_2 cos \theta$$. The only thing that could make it so you are getting the wrong answer is that $$\theta_2$$ is wrong...and that would mean that the diagram is wrong. Either that or the answer in the book is wrong.