Maximum kinetic friction-combined with a projectile problem

Click For Summary
SUMMARY

The discussion centers on a physics problem involving a cannon launching packets to a roof 125 meters high and 50 meters away. The initial speed required for the packets to land tangentially is calculated to be 9.9 m/s. The minimum coefficient of kinetic friction (μ) necessary to prevent the packets from sliding off the roof is determined to be approximately 0.1667, derived from the equation μ = a/g, where 'a' is the calculated acceleration of 1.6335 m/s² and 'g' is the acceleration due to gravity (9.8 m/s²).

PREREQUISITES
  • Understanding of projectile motion and kinematics
  • Familiarity with Newton's laws of motion
  • Knowledge of friction and its coefficients
  • Ability to solve quadratic equations in physics contexts
NEXT STEPS
  • Study projectile motion equations in-depth, focusing on horizontal and vertical components
  • Learn about the factors affecting friction, including surface materials and conditions
  • Explore advanced kinematics problems involving multiple forces
  • Investigate the use of simulation tools like Wolfram Alpha for solving physics problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of projectile motion and friction in real-world applications.

Poetria
Messages
267
Reaction score
42

Homework Statement


A cannon at ground level is shooting emergency packets to people stranded on the roof of a flooded building of height H=125 meters. The corner of the building is located a distance D=50 meters from the cannon. It is desired that the incoming packets are flying tangent to the roof as shown so that they land gently with as little impact as possible and slide along to a stop.

Find the initial speed v0 and at what angle θ (in degrees) the cannon should be aimed to achieve the above scenario
If the roof has length L=30 meters (not shown) what is the minimum coefficient of kinetic friction between the food packets and the roof such that the packets do not slide off the other side?

Homework Equations



v^2(x)=v^2_i,x+2a*x(x−xi)

The Attempt at a Solution



I only have a problem with the third part but I think I have found my mistake. I have confused horizontal velocity with vertical velocity. I got a very strange value for mu, much too high.

When packets land on a roof, there is no vertical velocity involved and I have already computed the value for the horizontal: v=9.9.


0=9.9^2-2*a*30
a=1.6335

m*a=mu*m*g
a=mu*g
mu=1.6335/9.8
mu=0.1666836734693877551020408163265306122448979591836734


 
Physics news on Phys.org
Looks correct.
Poetria said:
mu=0.1666836734693877551020408163265306122448979591836734
Are you sure it is not 0.1666836734693877551020408163265306122448979591836733[/color]? You should round final results to reasonable precision.
 
  • Like
Likes   Reactions: Poetria
Right. Many thanks. :)
It is my friend, Wolfram Alpha.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
40
Views
3K
Replies
2
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K