# Finding acceleration with Kinetic Friction coefficient

 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.
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P: 2,318
 Quote by phee 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.
Friction is miu * normal component of weight force.

EDIT: for the rest you are on your own as there is no detail of the set-up
 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=(F-29.4-37)/10 so if u have the force applied by the crane u can calculate the acceleration
P: 36
Finding acceleration with Kinetic Friction coefficient

 Quote by alphali 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=(F-29.4-37)/10 so if u have the force applied by the crane u can calculate the acceleration
Would the crane have applied a force? It isn't given in any of the text. All that is happening is that the board that is being held by the crane is tilting on a 22 degree angle causing the boxes to slide
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P: 2,318
 Quote by phee Would the crane have applied a force? It isn't given in any of the text. All that is happening is that the board that is being held by the crane is tilting on a 22 degree angle causing the boxes to slide
We need a picture !!! ?

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?
P: 36
 Quote by PeterO We need a picture !!! ? 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?
It only says,

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?
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P: 2,318
 Quote by phee It only says, 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?
With the plank at 22 degrees, you can find the component of the weight of a box parallel to, and perpendicular to, the plank.
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.
 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.
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P: 2,318
 Quote by phee 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.
Fkinetic friction = Uk*Fn
= (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
 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
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P: 2,318
 Quote by phee 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
It all looks fine [hoping your sin and cos values were correct, I didn't check them but they look reasonable].

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