# Solving Mechanics: Friction of Sledge on Slope

• MHB
• Shah 72
In summary, a boy drags a sledge of mass 4kg from rest down a rough slope at an angle of 18 degrees to the horizontal. He pulls it with a force of 8N for 3 seconds by a rope that is angled at 10 degrees above the parallel down the slope. After 3 seconds, the rope becomes detached from the sledge. The coefficient of friction between the slope and the sledge is 0.4. Using the equations of motion, it was found that the total displacement of the sledge down the slope was 18.5 meters. The normal reaction force and the force between the slope and the sledge were not included in the calculations.
Shah 72
MHB
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

during the drag ...

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

try again

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

recheck your acceleration calculation during the drag …

skeeter said:
recheck your acceleration calculation during the drag …

https://www.physicsforums.com/attachments/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)

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

skeeter said:
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!

skeeter said:
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?

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,

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

Last edited by a moderator:
skeeter said:
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!
skeeter said:
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!

## 1. What is friction and how does it affect a sledge on a slope?

Friction is the force that resists the motion of an object when it is in contact with another surface. When a sledge is on a slope, friction acts in the opposite direction of the sledge's motion, making it more difficult for the sledge to slide down the slope.

## 2. How does the angle of the slope affect the friction of a sledge?

The steeper the slope, the greater the friction on the sledge. This is because the normal force, which is the force that pushes the sledge into the slope, increases as the slope angle increases. This increased normal force results in a greater friction force.

## 3. What factors influence the coefficient of friction between the sledge and the slope?

The coefficient of friction is influenced by the types of materials in contact and the roughness of their surfaces. In the case of a sledge on a slope, the coefficient of friction is affected by the materials of the sledge and the slope, as well as any external factors such as moisture or debris on the surface.

## 4. How can the friction of a sledge on a slope be calculated?

The friction force can be calculated using the equation Ff = μN, where Ff is the friction force, μ is the coefficient of friction, and N is the normal force. The normal force can be calculated by multiplying the mass of the sledge by the cosine of the slope angle.

## 5. How can friction be reduced to make it easier for a sledge to slide down a slope?

Friction can be reduced by using materials with lower coefficients of friction, such as adding a layer of lubricant between the sledge and the slope. Additionally, decreasing the slope angle can also reduce the friction force on the sledge. However, it is important to note that reducing friction may also decrease the control and stability of the sledge on the slope.

• General Math
Replies
9
Views
1K
• General Math
Replies
6
Views
1K
• General Math
Replies
8
Views
921
• General Math
Replies
1
Views
713
• General Math
Replies
2
Views
909
• General Math
Replies
4
Views
934
• General Math
Replies
5
Views
847
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
419
• Introductory Physics Homework Help
Replies
3
Views
1K