Nails going down the side of the roof

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Homework Help Overview

The problem involves calculating the initial speed required to push a box of nails down a sloped roof so that it comes to rest at the edge. The roof is pitched at an angle of 22 degrees, and the coefficient of kinetic friction is given. The scenario includes considerations of forces acting on the box, including gravity and friction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of forces acting on the box, including gravity and friction, and how these forces affect acceleration. There are attempts to clarify the correct use of trigonometric functions in the context of the problem.

Discussion Status

Some participants are exploring different interpretations of the problem setup, particularly regarding the distance and direction involved. There is ongoing dialogue about the calculations and the signs of the forces, with no clear consensus reached yet.

Contextual Notes

Participants express confusion over the assumptions made regarding the geometry of the problem, particularly the interpretation of the distance of 5 meters and its relation to the angle of the roof. There are indications that the calculations may be affected by the choice of sine and cosine functions in determining forces.

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Homework Statement


You and your friend Peter are putting new shingles on a roof pitched at 22 degree . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.0 m away, asks you for the box of nails. Rather than carry the 2.5 kg box of nails down to Peter, you decide to give the box a push and have it slide down to him.

If the coefficient of kinetic friction between the box and the roof is 0.53, with what speed should you push the box to have it gently come to rest right at the edge of the roof?


Homework Equations



Vf^2=Vi^2+2*a*delta s

F=ma

The Attempt at a Solution



First found the length of the roof by assuming that the height was 5m and angle 22* i found the length to be 13.3475meters using trig...sin(22)=5m/roof length

then i did the force calculations to find Fnet then find the acceleration:

9.8cos(22)*2.5kg=22.71N

9.8cos(22)*2.5kg*.53= -12.039N

Fnet=22.71-12.039N
Fnet=10.671N

Then solved for the acceleration:

F=ma

10.671N=2.5kg * a
a=4.2684m/s^2

then i plugged those values into this equation to find Vi

Vf^2=Vi^2+2*a*delta s

0=Vi^2 + 2 * (4.2684m/s^2) * (13.3475m)

Vi=10.67m/s

is this correct? thank you!
 
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For the net force, you calculated both gravity's force and the frictional force using cosine. The force of gravity down the ramp should be sine.
 
I did both with sin and cosin and they were both wrong need help someone please help!
 
well i got 1.144856m/s^2 after getting the net force along the roofs axis to be 2.862N and found the Vi to be 5.53m/s and that is also incorrect is it suppose to be negative?

Im having a lot of trouble here...
 
Well the acceleration should be negative because the frictional force exceeds the force due to gravity.

Maybe your problem is in assuming "5 meters away" meant directly away from the one on top of the roof rather than down the roof.
 

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