Nails going down the side of the roof

  • Thread starter Thread starter charan1
  • Start date Start date
AI Thread Summary
The discussion centers on calculating the initial speed required to push a 2.5 kg box of nails down a 22-degree pitched roof so it comes to rest at the edge. The participant initially calculated the roof length and forces but struggled with the correct application of trigonometric functions for gravitational force, mistakenly using cosine instead of sine. After recalculating, they found the net force and acceleration, but still questioned the correctness of their results, particularly regarding the direction of acceleration and the interpretation of the distance measurement. Clarifications were provided that the frictional force should be accounted for as negative, and the distance should be understood as along the roof's slope. The conversation highlights the importance of proper force analysis in physics problems involving inclined planes.
charan1
Messages
40
Reaction score
0

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!
 
Physics news on Phys.org
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top