How Do You Calculate Friction Coefficients in Physics Without Mass?

[jen333]

Hey,
i have two 2D vector problems here that I'm stuck on! help is appreciated!

1) A boy sits at the top of a slide. When the slide makes an angle of 35.0 degrees with the horizontal, the boy is just on the verge of slipping. If the boy starts to move down the slide, the slide angle can be reduced to 33.0 degrees and the boy is then able to descend the slope at 2.00m/s. Find µk and µs. (the answers, respectively, are 0.649 and 0.700)

I've drawn diagrams of the problem already, but what I'm stuck on is just how to calculate µs and µk. all of the formulas that i am aware of calculating these use mass. ie) µk= Ff/Fn. what sort of ways can i execute this problem without using mass?


2) A 45.4kg box is placed at the top of an incline (µs = 0.500, µk= 0.400) will the box move? If it does, what is its speed at the bottom of the incline? (the answer is 4.25m/s)

i'm not sure, but I'm thinking that there will have to be an acceleration greater than 9.81m/s^2 inorder to move. i have a diagram here in which a 45.4kg box is placed on an incline at 30.0 degrees to the horizontal and is 3.00m high.

 

[OlderDan]

Hey,
i have two 2D vector problems here that I'm stuck on! help is appreciated!

1) A boy sits at the top of a slide. When the slide makes an angle of 35.0 degrees with the horizontal, the boy is just on the verge of slipping. If the boy starts to move down the slide, the slide angle can be reduced to 33.0 degrees and the boy is then able to descend the slope at 2.00m/s. Find µk and µs. (the answers, respectively, are 0.649 and 0.700)

I've drawn diagrams of the problem already, but what I'm stuck on is just how to calculate µs and µk. all of the formulas that i am aware of calculating these use mass. ie) µk= Ff/Fn. what sort of ways can i execute this problem without using mass?


2) A 45.4kg box is placed at the top of an incline (µs = 0.500, µk= 0.400) will the box move? If it does, what is its speed at the bottom of the incline? (the answer is 4.25m/s)

i'm not sure, but I'm thinking that there will have to be an acceleration greater than 9.81m/s^2 inorder to move. i have a diagram here in which a 45.4kg box is placed on an incline at 30.0 degrees to the horizontal and is 3.00m high.

1) Gravity pulls the boy straight downward, You can break the weight vector into two components; one is parallel to the slde and the other is perpendiculat to the slide. The component of weight that is perpendicular to the slide must be equal and opposite the normal force. The component of weith that is parallel to the slide must be equal and opposite the friction. Every force in this problem will be proporional to the boy's mass. It divides out of the equations.

After you get #1, #2 will make more sense. In this case friction will not be able to offset the component of weight parallel to the incline. There will be a resulting acceleration of the mass, but it will be far less than g.

 

[thepatientmental]

For the first problem, to calculate µk and µs, you can use the fact that at the point where the boy is just on the verge of slipping, the frictional force (Ff) is equal to the maximum static frictional force (µsFn). This means that µs = Ff/Fn. You can also use the fact that at the point where the boy is descending the slope at a constant speed, the frictional force (Ff) is equal to the kinetic frictional force (µkFn). This means that µk = Ff/Fn. From there, you can use the given information about the slide angle and the boy's speed to solve for µk and µs.

For the second problem, you are correct in thinking that there needs to be an acceleration greater than 9.81m/s^2 for the box to move. To determine if the box will move, you can use the equation Ff = µsFn to calculate the maximum static frictional force. If this force is greater than the force of gravity on the box (mg), then the box will not move. However, if the force of gravity is greater than the maximum static frictional force, then the box will start to slide down the incline. To calculate the speed of the box at the bottom of the incline, you can use the equation v = √(2gh), where h is the height of the incline. This will give you the final speed of the box at the bottom of the incline.