UFPM Challenge - Atwood Machine

UFPM Challenge!

Atwood Machine Experiment

Mission: Land the mass from the modified Atwood Machine on the constant velocity cart.

We started off by finding the total weight of the system: 

The weight of the cart + the 300 g added = 838.1 g, the total weight of the system.

We then converted grams to newtons and created a free-body diagram to better visualize the forces acting upon the Atwood Machine. 

838.1/1000 = .8381 kg x 10 = 8.381 N

 

If we use the formula a = fnet/m, we can find ft as well as the acceleration because fnet = ft since ft is the only force moving the cart. 

a = .507 N/.8381 kg

a = .605 m/s^2

ft = 50.7 g/1000 = .0507 kg x 10 = .507 N

Next, we took data on the constant velocity cart to see exactly how fast the cart travels.

We found:

Sec  |  M

0       .22

1       .53        

2       .84

3       1.16

4       1.47

5       1.83

Using this information, we were able to calculate the constant velocity: .30 m/s^2

v = change in x/t

.30 = .53 - .22

(.31) .30 = .31/t (.31)

t = .093 m/s^2

Vertical distance = 81 cm

Vi = 0 cm

Change in position = 81 cm/100 = .81 cm

Using another kinematics formula, we can find t.

Change in x = 1/2a (change in t) + Vi(change in t)

.81 = 1/2(.605)t^2 + 0(change in t)

.81/.3025 = .0325t^2/.0325

Take the square root of 2.67768595 = The square root of t

t = 1.64 s

So far, we know:

a = .605 m/s^2

Change in position = .81 cm

t = 1.64 s

Vi = 0

Since t shows the amount of time it will take the cart on the ground to travel .81 m, we multiplied 1.64 by .30, the velocity, and found that it will take .492 m for the cart to align with the dropping weight!