How Fast Should You Push the Box of Nails Down the Roof?

[pdonovan]

Question:
You and your friend Peter are putting new shingles on a roof pitched at 25 . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.5 away, asks you for the box of nails. Rather than carry the 2.0 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 0.55, with what speed should you push the box to have it gently come to rest right at the edge of the roof?


My attempt:
I drew a free body diagram for the box while it is in motion. From there I concluded that:
Fnet = Fgsin20 - f
--> mgsin20 - .55N = ma
--> 2(9.8)sin20 - .55(2*9.8*cos20) = 2a
--> a = -1.7131

I then used the kinematics equations Vf = Vi + aT; and Sf = ViT + .5a(T^2) to find Vi = 4.34m/s.

Any help or guidance is greatly appreciated! Thank you.

 

[ehild]

The method is fine, but the roof is inclined at 25 degrees to the horizontal, is not it? Why did you use 20 degrees?

ehild

 

[RTW69]

Are we using the correct terminology for the roof slope in this problem? The pitch of a roof is the rise in a 12 inch run. A 12 pitch rises 12 inches and runs 12 inches for a 45 degree angle. A 25 pitch rises 25 inches in a 12 inch run for a slope of 64 degrees. When they use the term pitch in this problem do they mean the slope is 25 degrees or how a carpenter would use the term?

 

[ehild]

I thought it was angle, otherwise why did Pdonovan use sine and cosine? I am almost sure a carpenter would use it on the other way... But the pitch was given as 25, it can not correspond to a 20° incline.

ehild

http://chestofbooks.com/architecture/Cyclopedia-Carpentry-Building-1-3/Pitch-Of-Roof.html

 

[rwisz]

Your approach and calculations are correct. To find the initial velocity, you used the equation Vf = Vi + aT, where Vf is the final velocity (which we want to be zero), Vi is the initial velocity, a is the acceleration (which you correctly calculated as -1.7131 m/s^2), and T is the time it takes for the box to reach the edge of the roof. However, in this case, we do not know the time T, so we need to use another equation to find it.

To find the time, we can use the equation Sf = ViT + .5a(T^2), where Sf is the final displacement (which we want to be 5.5 m), Vi is the initial velocity (which we are solving for), a is the acceleration, and T is the time. Plugging in the values, we get:

5.5 = 4.34T + .5(-1.7131)(T^2)
Simplifying, we get a quadratic equation: -0.8565T^2 + 4.34T - 5.5 = 0

Solving for T using the quadratic formula, we get T = 1.86 seconds.

Now, to find the initial velocity, we can plug in the value of T into the equation Vf = Vi + aT, and we get:

0 = Vi + (-1.7131)(1.86)
Solving for Vi, we get Vi = 3.19 m/s.

Therefore, to have the box gently come to rest at the edge of the roof, you should push it with an initial velocity of 3.19 m/s. Great job with your calculations!