- 957

- 0

**1. Homework Statement**

There must be a thousand variations on this but my step daughter brought this to me after i volunteered to help her with her Physics I class.

Burglar wants safe--decides to ice the floor thinking this will help him push it out the door. Coefficient of friction same for buglar and safe, both static and kinetic. Applies a fixed force at 30 degrees off vertical. I've change the constants provided, as I just want to make sure I understand principles.

Mass(burglar)=60Kg Mass Safe=600Kg Force applied=1200N MUs=0.2 MUk=.15

Does safe move, burglar or both, if so at what rate(s) of acceleration? If not what force is required at this angle to move the safe?

**2. Homework Equations**

F=ma, Frictional F=Normal*(mu)

**3. The Attempt at a Solution**

My thinking was that the force applied somewhat vertically helped in his efforts--it lightened the safe, reducing its Frictional force, while the reactive component helped dig in his heels (cramp-ons were not allowed). So I first computed the frictional forces.

If I'm thinking about this correctly,

frictional force for the burglar is his weight plus the vertical component of the reactive force, times Mu,

Making g=10m/s^2

that would be (60*10 + cos(30)*1200))*0.2=327.8

for the safe ((600*10 minus cos(30)*1200))*0.2=992.2

Since the horizontal component of his force is sin(30)*1200=600

it would not be enough to overcome the static friction of the safe, yet is larger than his frictional force, so he would accelerate backwards while the safe wouldn't budge. And that would just be the two differences in the forces (after adjusting the frictional force with the kinetic MU)/mass.

In order to get the safe to move:

6000*0.2/(.2*cos(30)+sin(30))=1782 N

At which point his frictional force will have increased to 482, which is still less than the reactive horizontal force, so he will still go backwards. And that the only way for this not to happen is to make the push more vertical still. This just doesn't seem intuitive.

Am I all wet here?

JS

Last edited: