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joemost12

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## Homework Statement

100 kg box is held in place on a ramp that rises at 30° above the horizontal. There is a massless rope joint to the box that makes a 22° angle above the surface of the ramp. Coefficients of friction between the box and the surface of the ramp are μk = 0.40 and μs = .60. The pulley has a mass of 10 kg and a radius of 10 cm. However the pulley does not have a friction.

A). What is the Max weight the hanging mass can have so that the box remains at rest.

B). The system is at rest with the weight of the hanging mass found in Part A, but a speck of dust comes to rest on the weight on the hanging mass, making the system unstable. What is the acceleration of the box at the very moment it starts moving?See diagram

https://i.imgur.com/xCWIHWb.png

## Homework Equations

## The Attempt at a Solution

So I think I got part A, and sort of got part B setup.. but I am not really sure where to go from there. [/B]

**Part A)**

Here is how I setup my equations:

Force Gravity on Block = (100)(9.8)sin(30) = 490

NF = 100cos(30) - Tension*sin(22) = 50√3 - Tension*sin(22)

Mass on Ramp =

Net force on Box = 0, forces equation =

Tension*cos(22) - M

I then proceeded to plug in

Then using the equation above and the Hanging mass equation I solved for

Here is how I setup my equations:

Force Gravity on Block = (100)(9.8)sin(30) = 490

NF = 100cos(30) - Tension*sin(22) = 50√3 - Tension*sin(22)

Mass on Ramp =

**M**

MHanging mass = Tension-(_{a}M

_{b}=**M**)(9.8) = 0_{b}Net force on Box = 0, forces equation =

Tension*cos(22) - M

_{a}gsin(30) - (**M**gcos(30) - Tension*sin(22) )μ = 0_{a}I then proceeded to plug in

**M**_{a}, g = 9.8, and μ = .60Then using the equation above and the Hanging mass equation I solved for

**M**_{b}I got

**M**

_{b}= ~ 88 kg**Part B)**

I am fairly confident about my answer of Part A.. but now unsure exactly how to solve part B.

I know the radius of the pulley is 10 cm and the mass is 10 kg.

I am fairly confident about my answer of Part A.. but now unsure exactly how to solve part B.

I know the radius of the pulley is 10 cm and the mass is 10 kg.

I had to find the equations for the two tensions. However, one thing I had to form a new equation for (at least I think ) is the Normal Force. Before I knew the vertical component of the normal force would just be 0, but now since it is moving it could also change.

My new equation is:

**NF = 100cos(30) - Tension*sin(22) =**

I know I need to substitute this into my original equation, so I can use it to calculate the force of Friction

Tension2*cos(22) - M

put this equation in terms of Tension2, but I must leave acceleration in terms of A

Tension2 = 129A + 770

My equation for tension1 was easier, just:

Tension1 =

Now I plug them into a Torque equation taking into account the mass of the pulley and radius

Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝

Now here is where my struggles appear, I have two unknowns with A and

**M**

NF = 100cos(30) - Tension*sin(22) -_{a}ANF = 100cos(30) - Tension*sin(22) -

**M**_{a}A = 0I know I need to substitute this into my original equation, so I can use it to calculate the force of Friction

Tension2*cos(22) - M

_{a}gsin(30) - (**100gcos(30) - Tension2*sin(22) - M**)μ = 0_{a}Aput this equation in terms of Tension2, but I must leave acceleration in terms of A

Tension2 = 129A + 770

My equation for tension1 was easier, just:

Tension1 =

**M**A -_{b}**M**_{b}ANow I plug them into a Torque equation taking into account the mass of the pulley and radius

Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝

Now here is where my struggles appear, I have two unknowns with A and

**M**Let me know if there is anything I should elaborate on something.. ,thanks in advance!_{b}. I've got to imagine that the intention of Mass B is that the speck of dust would represent just about nothing, and the point of it was to explain that the system would be in motion and I should just account for it as the same mass from before, but I don't know. ANy advice on this would be great!
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