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Physics Mechanics- friction

Shah 72

Member
Apr 14, 2021
193
A boy drags a sledge of mass 4kg from rest down a rough slope at an angle of 18 degree to the horizontal. He pulls it with a force of 8N for 3s by a rope that is angled at 10 degree above the parallel down the slope. After 3s the rope becomes detached from the sledge. The coefficient of friction between the slope and the sledge is 0.4. Find the total distance the sledge has moved down the slope from when the boy started dragging it until it comes to rest.
For the first part before the rope is detached
I calculated
R=40cos 18+8sin 10
Coefficient of friction =0.514
F=m×a
So I got a = 0.003m/s^2.
I then calculated s= ut +1/2 at^2 and got s1 =0.0135m.
I calculated v=0.009m/s
Part 2 when the rope is detached
R=40cos 18
By using F=m×a
I got a=0.714m/s^2.
Iam not getting the right ans for the total distance.
The ans in the textbook is 18.5m.
Pls help
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
during the drag ...

$ma = mg\sin(18) + 8\cos(10) - \mu[mg\cos(18) - 8\sin(10)]$

try again
 

Shah 72

Member
Apr 14, 2021
193
I did your way and got a= 0.35m/s^2, v=1.05m/s and s=1.575m
Now the second part when the rope is detached
R=40cos18, F=40sin18 -0.4×40cos18, F=2.84N, using F=m×a, I get a=0.71m/s^2, and s2=0.776m. I still don't get the ans 18.5m
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
recheck your acceleration calculation during the drag …

D90B437C-D771-43CE-A3C4-230F944E72E3.jpeg
 

Shah 72

Member
Apr 14, 2021
193
recheck your acceleration calculation during the drag …

View attachment 11134
I did not understand in the second part when the rope is detached, how is it 10 sin 18 + 2 cos 10. Shouldn't it be R = 40 cos 18 and F= 40sin 18- 0.4(40cos 18)
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
during the drag …

$a_1 = 10\sin(18) + 2\cos(10) - 0.4[10\cos(18) - 2\sin(10)] = 1.39 \, m/s^2$

$\Delta x_1 = \dfrac{1}{2}a_1 \cdot 3^2 = 6.28 \, m$

$v_1 = 0 + a_1 \cdot 3 = 4.18 \, m/s$


after the rope is detached …

$a_2 = 10\sin(18) - 4\cos(18) = -0.714 \, m/s^2$

$\Delta x_2 = \dfrac{0^2- v_1^2}{2a_2} = 12.25 \, m$

total displacement = 6.28 + 12.25 = 18.5 m
 

Shah 72

Member
Apr 14, 2021
193
during the drag …

$a_1 = 10\sin(18) + 2\cos(10) - 0.4[10\cos(18) - 2\sin(10)] = 1.39 \, m/s^2$

$\Delta x_1 = \dfrac{1}{2}a_1 \cdot 3^2 = 6.28 \, m$

$v_1 = 0 + a_1 \cdot 3 = 4.18 \, m/s$


after the rope is detached …

$a_2 = 10\sin(18) - 4\cos(18) = -0.714 \, m/s^2$

$\Delta x_2 = \dfrac{0^2- v_1^2}{2a_2} = 12.25 \, m$

total displacement = 6.28 + 12.25 = 18.5 m
Thank you very much!!!
 

Shah 72

Member
Apr 14, 2021
193
during the drag …

$a_1 = 10\sin(18) + 2\cos(10) - 0.4[10\cos(18) - 2\sin(10)] = 1.39 \, m/s^2$

$\Delta x_1 = \dfrac{1}{2}a_1 \cdot 3^2 = 6.28 \, m$

$v_1 = 0 + a_1 \cdot 3 = 4.18 \, m/s$


after the rope is detached …

$a_2 = 10\sin(18) - 4\cos(18) = -0.714 \, m/s^2$

$\Delta x_2 = \dfrac{0^2- v_1^2}{2a_2} = 12.25 \, m$

total displacement = 6.28 + 12.25 = 18.5 m
Can you also pls explain why I don't need to take the normal contact force and the force between the slope and the sledge?
20210510_100241.jpg
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
935
The Normal Reaction force, $N$, is perpendicular to the incline. Reference the two FBD's attached ...

during the drag by the 8N applied force,

$N_1 = mg\cos{\theta} - F\sin{\phi} \implies f_1 = \mu(mg\cos{\theta} - F\sin{\phi})$

after the dragging force is gone,

$N_2 = mg\cos{\theta} \implies f_2 = \mu mg\cos{\theta}$

incline_FBDx2.jpg
 
Last edited:

Shah 72

Member
Apr 14, 2021
193
during the drag …

$a_1 = 10\sin(18) + 2\cos(10) - 0.4[10\cos(18) - 2\sin(10)] = 1.39 \, m/s^2$

$\Delta x_1 = \dfrac{1}{2}a_1 \cdot 3^2 = 6.28 \, m$

$v_1 = 0 + a_1 \cdot 3 = 4.18 \, m/s$


after the rope is detached …

$a_2 = 10\sin(18) - 4\cos(18) = -0.714 \, m/s^2$

$\Delta x_2 = \dfrac{0^2- v_1^2}{2a_2} = 12.25 \, m$

total displacement = 6.28 + 12.25 = 18.5 m
I understood. Thank you so so so much!!!
The Normal Reaction force, $N$, is perpendicular to the incline. Reference the two FBD's attached ...

during the drag by the 8N applied force,

$N_1 = mg\cos{\theta} - F\sin{\phi} \implies f_1 = \mu(mg\cos{\theta} - F\sin{\phi})$

after the dragging force is gone,

$N_2 = mg\cos{\theta} \implies f_2 = \mu mg\cos{\theta}$

View attachment 11136

View attachment 11136
I understood this. I was doing the silly mistake of finding a new coefficient of friction. Thank you so so much for explaining!!!