What Incline Is Needed for a Toboggan to Slide Downhill?

[kamalpreet122]

Homework Statement

On a cold day, a 35kg child takes a 3.5 kg toboggan to a snowy hill to slide down it. There is U_s = 0.15 and U_k=0.05 between the hill and the toboggan. What is the minimum incline required before the toboggan will slide? What angle is required for the child to accelerate at half the rate of gravity?

The Attempt at a Solution

Given:
Us (coefficient of static friction) = 0.15
Uk (coefficient of kinematic friction) = 0.05

Ff (force of friction) = -X-component of Weight (mg)
Fn (normal force) = - Y-component of weight (mg)

therefore...
Ff = -mgCos[tex]\theta[/tex]-------- 1
Fn = -mgSin[tex]\theta[/tex]-------- 2

From equation 1

mg = - Ff / Cos[tex]\theta[/tex] ----- 3

Sub in 3 into 2

Fn = - (-Ff/Cos[tex]\theta[/tex] )( Sin[tex]\theta[/tex] )
Fn = FfTan[tex]\theta[/tex]

Minimum angle required for the object to slide is Critical Angle.. (therefore theta now represents critical angle)

--> Fn= FfTan[tex]\theta[/tex] ------ 4

Ff = U_sFn ---------5

Sub in 5 into 4

Fn = U_sFnTan[tex]\theta[/tex]

Fn / U_sFn = Tan[tex]\theta[/tex]

Tan[tex]\theta[/tex] = 1 / U_s

Tan[tex]\theta[/tex] = 1 / 0.15

-----> Tan[tex]\theta[/tex] = 6.67
[tex]\theta[/tex] = Tan^-1(6.67)
[tex]\theta[/tex] = 81.5

So therefore 81.5 degrees is the minimum incline required for the toboggan to slide.. I think that's the answer to the first part of this problem.. I am not too sure though because I'm in grade 11 and haven't been taught much about such types of problems. Can someone please help me do the second part and check if i did the first part right.. i would really really appreciate it.

 

[amy andrews]

It looks like you've done the first part correctly; it's a somewhat unconventional approach but works just fine!
For the second part, you'll have to use the equation sum of forces= mass*acceleration. Recalculate the forces using the acceleration and mass given, the y-components of the forces will still add to zero so calculate the x-component of the force. Then work backwards to calculate theta.
Hope this helps.

 

[kamalpreet122]

i get that but the problem i face is .. there is no given acceleration =(

 

[kamalpreet122]

or do i use half of g (9.81/2) as the given acceleration ?

 

[amy andrews]

Yep, the given acceleration is just (9.8/2) m/s^2 or 4.9 m/s^2