Solving Ice's Acceleration Down a Sloped Roof

In summary, the problem involves a 1.5 kg chunk of ice sliding down a 60 degree sloped roof. The acceleration of the ice is 8.48 m/s/s with no friction, and with a friction force of 3.0 N, the acceleration is found using Newton's second law of motion and a free body diagram. The acceleration is equal to the net force divided by the mass.
  • #1
runningirl
118
0

Homework Statement



A 1.5 kg chunk of ice slides down a roof sloped at a 60 degree angle with the horizontal.
a)If there is no friction, determine the acceleration of the ice.
b) If the force of the friciton is 3.0 N, what is the acceleration of the ice?

Homework Equations



f=ma


The Attempt at a Solution



for part a:
9.8*1.8*sin(60)=12.73 N
12.73/1.5=8.48 m/s/s

part b?!
 
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  • #2
runningirl said:
part b?!
Have you tried Newton's second law of motion? (Hint: Make sure you find the net force, since there are multiple forces involved.)
 
  • #3
collinsmark said:
Have you tried Newton's second law of motion? (Hint: Make sure you find the net force, since there are multiple forces involved.)

3+F(cos60)?
i don't know the force being pushed on the slope to find the horizontal force ...
 
  • #4
runningirl said:
3+F(cos60)?
i don't know the force being pushed on the slope to find the horizontal force ...
Try drawing a block diagram. It will (as always) guide you to setting up the equation(s) correctly. How does mg fit into all of this?

[Edit: And feel free to refer back on your work for part a). It may bring you some insight.]
 
Last edited:
  • #5
Start by drawing a free body diagram this will help you to see what forces are acting and which way(you are going to break the force of mg into components). Afterwards, you can now use this diagram to see what forces are equal,or cancel each other out, and in your case which is greater than the other(force to overcome the friction force). Then use similar techniques as in part a) to aid you in getting your answer.
 
  • #6
Force by gravity in direction of slope = 9.81*1.5*sin60
Net force = 9.81*1.5*sin60 - 3.0

Net force = ma
a = (Net force)/m
 

What is the concept of acceleration when solving ice's acceleration down a sloped roof?

The concept of acceleration when solving ice's acceleration down a sloped roof is the rate at which the ice gains speed as it slides down the roof. It is measured in meters per second squared (m/s^2).

How is the acceleration of ice down a sloped roof calculated?

The acceleration of ice down a sloped roof can be calculated using the formula a = gsinθ, where "a" is the acceleration, "g" is the acceleration due to gravity (9.8 m/s^2), and "θ" is the angle of the slope.

What factors affect the acceleration of ice down a sloped roof?

The factors that affect the acceleration of ice down a sloped roof include the angle of the slope, the surface texture and material of the roof, and any external forces such as wind or friction.

How does the mass of the ice affect its acceleration down a sloped roof?

The mass of the ice does not affect its acceleration down a sloped roof. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the applied force and inversely proportional to its mass. Since the force of gravity is the same for all objects, the mass of the ice does not impact its acceleration down the slope.

How can the acceleration of ice down a sloped roof be used in real-life applications?

The concept of acceleration of ice down a sloped roof can be applied in real-life situations such as designing safe and efficient roofs, predicting the speed at which ice or snow will slide off a roof, and understanding the effects of different weather conditions on the acceleration of ice down a sloped roof.

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