Solve Pulley Problems: Friction & Inertia | Physics Help

[blackice552]

I have to problems that I'm having trouble with today I just can't seem to get my head around a lot of these physics concept so I'm finally coming here Plz help me. I've tried to do these problems and I just don't know what I'm doing.



1.We have two masses hanging from a pulley of inertia mp and constant frictional force f. One mass is 21.5g and the other is 39.8g. When data is taken, the slope of the 1/a vs. (m1 + m2) graph turns out to be 6.77 1/N. What is the value of the frictional force of the pulley? (2 decimals and UNITS!) (g=9.8m/s2)

We have two masses hanging from a pulley of inertia mp and constant frictional force f. One mass is 55.9g and the other is 41.8g. When data is taken, the intercept of the 1/a vs. (m1 + m2) graph turns out to be 0.04s2/m and the frictional force of the pulley 0.31N. What is the value of the inertia of the pulley? (2 decimals and SI UNITS!) (g=9.8m/s2) (If you get a negative answer type it in even though it doesn't make sense physically)

Homework Equations

These are the equations I have:

for no friction:
1/a = (m1 + m2)/((m1 -m2)g)

for a system with friction:
(m1 - m2)g - f = (m1 + m2 + mp)a

1/a = [(m1 + m2)/((m1-m2)g - f)] + [mp/((m1-m2)g -f))]

The Attempt at a Solution

My solutions were totally outrageous I have no idea what I'm doing frankly I've studied this stuff for hours and can't wrap my head around it

 

[Nate810]

.For the first problem:1/a = (21.5 + 39.8)/((21.5 - 39.8)9.8)f = (21.5 - 39.8)9.8 - (21.5 + 39.8)6.77 f = -3.45 NFor the second problem:(55.9 - 41.8)9.8 - f = (55.9 + 41.8 + mp)0.04mp = ((55.9 - 41.8)9.8 - f)(25.1/0.04) - (55.9 + 41.8)mp = 0.31 N * 25.1/0.04 - 97.7mp = 7.60 kg

 

[m2003]

so I hope that someone can help me with these problems.

Dear student,

I understand that physics concepts can be difficult to grasp, but don't worry, I am here to help you solve these pulley problems. Let's take a step-by-step approach to solving them.

1. First, let's understand the given information. We have two masses hanging from a pulley with an inertia of mp and a constant frictional force of f. The masses are 21.5g and 39.8g. We are given the slope of the 1/a vs. (m1 + m2) graph, which is 6.77 1/N. We are asked to find the value of the frictional force of the pulley.

2. To solve this problem, we will use the equation for a system with friction: (m1 - m2)g - f = (m1 + m2 + mp)a. We can rearrange this equation to solve for the frictional force (f): f = (m1 - m2)g - (m1 + m2 + mp)a.

3. Now, let's plug in the given values into this equation. Remember to convert the masses from grams to kilograms (1g = 0.001kg) and the acceleration due to gravity (g) is 9.8m/s^2. We get: f = (0.0215kg - 0.0398kg) * 9.8m/s^2 - (0.0215kg + 0.0398kg + mp) * (6.77 1/N).

4. We are also given the value of the slope (6.77 1/N) and we know that the slope is equal to the inverse of the acceleration (a). Therefore, we can substitute the value of the slope into our equation as a = 1/6.77 1/N.

5. Now, we have all the necessary values to solve for the frictional force (f). We get: f = -0.00084kg * 9.8m/s^2 - (0.0613kg + mp) * (0.148 1/N).

6. To get the value of the frictional force in Newtons, we need to multiply it by 1/N. Therefore, our final equation becomes: f = -0.00084kg *