# Physics review

1. Dec 22, 2004

### mutzy188

My physics teacher gave us a huge packet with some review problems that we should know how to do for the test. There are 3 that I have no clue how to do.

1.) A morbidly obese man weighing 500 newtons stands 2.5 meters from the inner edge of a nail supporting a wooden platform. The wooden platform he stands on weighs 200 newtons, and is a total distance of 6 meters long. The platform is supported at the end by a steel cable (in blue) making a 45˚ angle above the horizontal.
Part A: Draw a free-body diagram representing this situation.
Part B: Find the tension T in the string (indicated by the blue arrow).
Part C: Find the force P that the nail exerts on the wooden platform (indicated by the green arrow). Give the magnitude of the force, along with the unknown angle theta.

I can do part B but that's about it. How do you do part a and c???

2.) A ball of mass M is attached to a string of length L and is swung in a vertical circle just above the ground. The tension T in the string at the top of its swing is 2g. The ball barely clears the ground at its minimum point.
Part A: Find the velocity of the ball at the top of its orbital. Express your answer only in terms of M, g, and L.
Part B: Find the tension T in the string at the bottom of its orbital. Express your answer only in terms of M, g, and L.
Part C: Find the centripetal acceleration of the ball at the top of its orbital. Express your answer only in terms of M, g, and L.

I don't know what to do. I know a= (v^2)/r but I don't have a. Is this the correct formula for this problem?

3.) A large block of mass 1.5 kg is attached to a spring, which is connected over a massless, frictionless pulley. Both of them are hung on opposite sides of the pulley, as illustrated in the diagram above. The spring has a force constant of 500 N/m, and is initially at its equilibrium position. The block is also released from its equilibrium position. After release, the block falls 3 centimeters before it hits the ground. Find the velocity v, in meters per second, just before the instant that the block strikes the ground.

I don't know what to do.

Any help would help. Thanks

2. Dec 23, 2004

### Diane_

1) a) In a free-body diagram, the object of interest is represented as a point. You then draw arrows to indicate the forces acting on it. Properly speaking, the arrows should point in the right direction and their length should be proportional to the magnitude of the force involved, but most people just get the length and direction "sorta" right.

c) Assume that the platform is static. What does that tell you about the net force? Using that, break the forces into components (horizontal and vertical will probably work well here) and resolve them accordingly.

2) You might want to check the problem again. It states that the tension is "2g". That might be an acceleration, but tension is a force. There's something odd there.

In any event, remember that the tension in the string at the top is providing some of the centripetal force. Where will the rest of it come from? Given that, what is the total centripetal force acting on the mass, M? You know the equation for centripetal acceleration - it should be easy to find the equation for centripetal force. (Remember Newton's Second Law.) Set your answer to that equation and start solving.

Is that enough to get you going?

3) The diagram is a little unclear, but: do a free body diagram of the block. Remember that the tension in the string is one of the forces acting on the block. You'll also want to do a free body diagram of the spring. If we assume that the string is massless and non-extensible, then it will simply transmit the same force along it's length, i.e. however hard it's pulling at one end is how hard it's pulling at the other. You will probably end up with a set of simultaneous equations, which you can then use to solve for the motion of the system.

Let me know if you need more.

3. Dec 23, 2004

### futb0l

mg = T = 2g
so m = 2 ?

4. Dec 23, 2004

### mutzy188

I don't get how to do 1 b.
(2.5)(500) + (6)(200)=0????????????

5. Dec 23, 2004

### mutzy188

1. C. the horizontal component is tcos theta but what is the other one??

2.) a. would the answer be v= sqrt( [r(2g+fg)]/m)

3.) mgh + .5mv^2 + .5kx^2
(9.8)(3)(1.5) + .5(1.5)v^2 + .5(500)(3^2)
v= 3058.8????????

6. Dec 23, 2004

### mutzy188

1. c.
t sin theta = 6
tcos theta = 2.5

(6^2) + (2.5^2)= t^2
t=6.5??

7. Dec 23, 2004

### mutzy188

im sorry. im rushing like this because i need 2 know this in like an hour. any help would be appreciated. thanks