Physics homework question (Dynamics)

[Matt1234]

Homework Statement

A rope exerts a force of magnitude 21 N, at an angle 31 deg above the horizontal, on a box at rest on a horizontal floor. The coefficients of friction between the box and floor are Us = 0.55 and Uk = 0.50. The box remains at rest. Determine the smallest possible mass of the box.

Homework Equations

U= Ff / Fn

The Attempt at a Solution

I don't know how to go about it yet, Sorry.

So far i drew a FBD which i believe is correct.

Any help with themethod would be appreciated.

 

[Matt1234]

nevermind i got it.

below is the solution.

Just a heads up, throughout the next few weeks i will be posting questions to some problems i have. I am pretty much using this site as my tutor since i cannot afford one. I appreciate all the help and thank you for being here.

 

[RoyalCat]

For any object subject to static friction the following holds true:
f_s < f_{s_{max}}=\mu _s N
For the critically small mass, assume that the mass is on the verge of slipping and starting to move. The FBD should help you analyze all the forces at work and extract m from the known variables.

Aww, a minute too late.

If you want to have your scans show up a bit quicker so people can respond to them before a moderator comes on, try uploading them to an image-sharing website and posting a link instead of posting them as attachments.

 

[Matt1234]

Will do in the furture, thank you.

I actually got a new problem. :)
Here it is:

A gymnast of mass 72Kg, initially hanging at rest from a bar. Let's go of the bar and falls vertically 0.92 m to the floor below. Upon landing, bends his knees, bringing himself to rest over a distance of 0.35 m. The floor exerts a constant force on him as he slows down. determine (b) the magnitude of the force the floor exerts on him as he slows down.

I did part (a) Which was to calculate the the velocity before hitting the floor.
Vf= 4.25 m/s

Thanks, this one i have been on for about 10 min so far.

 

[Matt1234]

Thanks to the original poster.

Although its about 50N off from the answer in my book it does make sense.

Since delta d = 0.35 m, g = -9.8 m/s^2, and Vf = (2g*0.92)^0.5, we have:
Vf = 4.24 m/s (speed as you hit the ground)

Vf^2 = Vo^2 + 2*a*d. This equation is valid for when you're slowing down/flexing your knees. In this equation, Vo = 4.24 from before and Vf = 0. Therefore we find that a = -25.8 m/s^2.

Since F = ma, F = 72kg * -25.8 m/s^2 = about 1850 N.

 

[PhanthomJay]

Thanks to the original poster.

Although its about 50N off from the answer in my book it does make sense.

Since delta d = 0.35 m, g = -9.8 m/s^2, and Vf = (2g*0.92)^0.5, we have:
Vf = 4.24 m/s (speed as you hit the ground)

Vf^2 = Vo^2 + 2*a*d. This equation is valid for when you're slowing down/flexing your knees. In this equation, Vo = 4.24 from before and Vf = 0. Therefore we find that a = -25.8 m/s^2.

Since F = ma, F = 72kg * -25.8 m/s^2 = about 1850 N.

Hi, Matt. You should post new problems in a separate post; otherwise, your post will tend to get lost in the abyss. The Force you have calculated , 1850 N, is the net force acting on the gymnast. That is not the force of the floor acting on the gymnast. Draw a FBD of the gymnast while he is in contact with the floor and identify all forces acting on him; then determine the floor force using Newton 2. I don't get anything close to your (or the book's ?) result.