Finding acceleration with Kinetic Friction coefficientby phee Tags: acceleration, coefficient, friction, kinetic 

#1
Aug1811, 06:51 PM

P: 36

1. The problem statement, all variables and given/known data
At a construction site, a small crane is raising two boxes of nails on a plank to the roof. One box has already been opened and is half full, while the other box is new. The boxes, including the nails, weigh 10kg (box 1) and 20kg (box 2), respectively and are both the same size. If the coefficient of kinetic friction is 0.3, how fast will the boxes accelerate along the plank, once they start to slide? This is for box 1: m=10kg Uk = 0.3 Weight = 10kg x 9.8 = 98 N angle = 22 2. Relevant equations Fnet = m/a Fgx = mg(sin22) 3. The attempt at a solution So so far I am trying to find the net force, so I concluded I needed to find Fgx Fgx = mg(sin22) = 37 N F(kinetic force) = 0.3(mg) = 29.4 I am not finished but was wondering if someone could at least hint to me as if I am going in the right direction. This has been confusing me for the past hour. 



#2
Aug1811, 06:58 PM

HW Helper
P: 2,316

EDIT: for the rest you are on your own as there is no detail of the setup 



#3
Aug1811, 09:06 PM

P: 24

Fnet=ma
F(force applied by the crane)Fr(force of friction)+w(force of gravity on the box)=ma projection along the direction of motion Wx=mgsin22 and Wx is in the opposite direction of motion so it will have negative value F(Uk.m.g)(mgsin22)=ma now i think ur missing the force applied by the crane in the givings because a=(F29.437)/10 so if u have the force applied by the crane u can calculate the acceleration 



#4
Aug1811, 09:08 PM

P: 36

Finding acceleration with Kinetic Friction coefficient 



#5
Aug1811, 11:41 PM

HW Helper
P: 2,316

You wondered if the crane would apply a force, because it wasn't given in the text. You are one step ahead of us as we don't even have the text!! EDIT: is the small crane lifting this all up at constant speed? 



#6
Aug2211, 05:55 PM

P: 36

a small crane is raising two boxes of nails on a plank to the roof. I assume its at a constant velocity which means no acceleration from the crane. The plank then tilts and the boxes begin to slide at a 22 degree angle. Do I find Fgx then plug it into Fnet = m/a? 



#7
Aug2211, 06:45 PM

HW Helper
P: 2,316

The perpendicular component [the normal force] will enable you to calculate the Friction force acting. Presumably the parallel component of weight exceeds this and you can then use F = ma to find the acceleration. Don't forget that the half empty box has smaller forces involved, but at the last step, a smaller mass. 



#8
Aug2211, 06:54 PM

P: 36

Let me spill what I have so far,
(for the 10kg box) Fgx = mg(sin22) Fgx = (10kg)(9.8 ms^2)(sin22) Fgx = 37 N Fn = Fgy Fgy= mg(cos22) Fgy= (10)(9.8)(cos22) Fgy= 91 N *Fn = 91 N* Fkinetic friction = Ukmg = (0.3)(10)(9.8) =29.4 N So... Fnet = 91 N  37 N  29.4 N Fnet = 24.6 N Fnet=ma a = 10 kg / 24.6 N a = 0.4 m/s ^2 I hope I got it, I am not 100% sure I should be using Fgy in this equation though. 



#9
Aug2211, 07:12 PM

HW Helper
P: 2,316

= (0.3) * 91 = what ever you get. Friction force is calculated from the NORMAL force, not the weight. Then Fnet = 37 N  (29.4 N) I left your old 29.4 here  it should be the new value you calculate using the normal force. The problem could be affected by the crane if the crane was accelerating the plank. In the absence of any indication otherwise, we can assume it is travelling at constant speed  the usual way cranes work 



#10
Aug2211, 07:20 PM

P: 36

Fnet = Fgx  Fn
= 37 N  27.3 N = 9.7 N Fnet = ma 9.7 N = (10kg)a 9.7 N /10kg = a 0.97 m/s ^2 = a Therefore the acceleration of the 10 kg box is 0.97 m/s ^ 2 This is right? And thank you for the help kind sir 



#11
Aug2211, 07:21 PM

HW Helper
P: 2,316




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