Inclined Plane Newton's Third Law

AI Thread Summary
To determine the coefficient of kinetic friction for a box sliding down a 39° ramp with an acceleration of 1.32 m/s², the net acceleration acting on the box must be analyzed. The gravitational acceleration component down the ramp is calculated as 9.8 m/s² multiplied by sin(39°), resulting in approximately 6.2 m/s². The frictional acceleration is found by subtracting the given acceleration from this value, yielding about 4.8 m/s². The normal force is derived from the gravitational component perpendicular to the ramp, calculated as 9.8 m/s² times cos(39°), which is approximately 7.6 m/s². The coefficient of kinetic friction is then determined by dividing the frictional acceleration by the normal acceleration, resulting in a value of approximately 0.64.
hshphyss
Messages
45
Reaction score
0
Can anyone help me with this problem? thank-you

A box slides down a 39° ramp with an acceleration of 1.32 m/s2. Determine the coefficient of kinetic friction between the box and the ramp
 
Physics news on Phys.org
Please show what you've done so far.
 
I don't know where to start because of the many unknwns. I know that mgcos39=N.
 
I m sorry, if its too late!

Just registered myself...so i don't know if the answer wud still help u!
in ur quest...the acceleration given is the net acceleration actin on the object that's slidin down the ramp. so we know that the difference between the acceleration actin horizontally down the ramp and the acceleration actin opposite to it(acceleration occurin in the direction of friction) has to be 1.32m/s/s.
all we need ta do now is to find the acceleration actin horizontally down the ramp. which has ta be 9.8 times sin39= 6.2m/s/s
as i already mentioned the frictional acceleration has ta be 1.32 less than 6.2(6.2-1.3=4.8m/s/s
so now we know the acceleration in the direction of kinetic friction. acceleration occurin in the direction of normal is balanced by the acceleration towards the gravity. so if we find the y component of gravity...we hav the normal(as it has ta be the same!
thats 9.8 cos39=7.6m/s/s
co-eff of friction= accl towrds friction/acceleration normal
4.8/7.6=.64
thats bout it i guess...generally we use force instead of acceleration...but in this prob we assume mass constant!
if u hav further doubts...u can always post it, it will help me learn more...
take care
 
wow thank-you so much
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top