Dynamics - the physics of tobogganing and forces at an angle

In summary, the conversation discusses a problem where a child is tobogganing down a hillside with a combined mass of 50.0kg. The slope is at an angle of 30.0° to the horizontal and the goal is to find the acceleration of the child in two scenarios: (a) when there is no friction and (b) when the coefficient of friction is 0.15. The solution involves calculating various forces such as the force of gravity, normal force, and force of kinetic friction. The final answers are +4.9m/s/s for no friction and +3.63m/s/s for a coefficient of friction of 0.15.
  • #1
L_0611
24
0

Homework Statement



A child is tobogganing down a hillside. The child and the toboggan together have a mass of 50.0kg. The slope is at an angle of 30.0° to the horizontal.
Assume that the positive y-direction is pointing in the direction of the normal force. Assume that the positive x-direction is down the incline.

Find the acceleration of the child

a) in the case where there is no friction

b) if the coefficient of friction is 0.15

The Attempt at a Solution



a) Fn + FgI = mgcosθ
= (50)(9.8)cos 30°
= 424N
Fnet = FgII = mgsinθ
= (50)(9.8)sin 30°
= 245 N (+ x-direction)
Fnet = ma
a = Fnet/m
= 245/50
= +4.9m/s/s
Therefor the child's acceleration in the case where there is no friction is 4.9 m/s/s.

b) FgI = 424 (- y-direction)
FgII = 245 (+ x-direction)

Fnet = FgII (+ x-direction) + Fk (- x-direction)

Fn = FgI = 424N (down)

Fk = μkFn
= (0.15)(424)
= 63.6N (- x-direction)

Fnet = FgII + Fk
= (+245)+(-63.6)
= +181.4N
Fnet =ma
a = Fnet/m
= 181.4/50
= +3.63m/s/s

Therefore the child's acceleration is +3.63 m/s/s if the coefficient of friction is 0.15.
 
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  • #2
L_0611 said:

Homework Statement



A child is tobogganing down a hillside. The child and the toboggan together have a mass of 50.0kg. The slope is at an angle of 30.0° to the horizontal.
Assume that the positive y-direction is pointing in the direction of the normal force. Assume that the positive x-direction is down the incline.

Find the acceleration of the child

a) in the case where there is no friction

b) if the coefficient of friction is 0.15

The Attempt at a Solution



a) Fn + FgI = mgcosθ
= (50)(9.8)cos 30°
= 424N
Fnet = FgII = mgsinθ
= (50)(9.8)sin 30°
= 245 N (+ x-direction)
Fnet = ma
a = Fnet/m
= 245/50
= +4.9m/s/s
Therefor the child's acceleration in the case where there is no friction is 4.9 m/s/s.

b) FgI = 424 (- y-direction)
FgII = 245 (+ x-direction)

Fnet = FgII (+ x-direction) + Fk (- x-direction)

Fn = FgI = 424N (down)

Fk = μkFn
= (0.15)(424)
= 63.6N (- x-direction)

Fnet = FgII + Fk
= (+245)+(-63.6)
= +181.4N
Fnet =ma
a = Fnet/m
= 181.4/50
= +3.63m/s/s

Therefore the child's acceleration is +3.63 m/s/s if the coefficient of friction is 0.15.

I think your answers are right. I'm having a hard time puzzling through your designations for the various forces though. Could you explain them?
 
  • #3
yeah I wasn't sure how to write them on the computer, basically I am following my book.
Fg - force of gravity
FgI - Fg(cosθ)
FgII = Fg(sinθ)
Fn - normal force
Fnet - net force
Fk - force of kinetic friction
m - mass
a - acceleration
g - gravity
μk - coefficient of kinetic friction

please let me know if I missed any or if you still have any questions. Thanks
 

Related to Dynamics - the physics of tobogganing and forces at an angle

1. What is the physics behind tobogganing?

The physics behind tobogganing involves a combination of several forces, including gravity, friction, and air resistance. When a toboggan is placed on an inclined surface, gravity pulls it downwards, while friction between the toboggan and the surface helps to slow it down. Air resistance also plays a role in slowing down the toboggan as it moves through the air.

2. How do forces at an angle affect tobogganing?

Forces at an angle, also known as vector forces, can affect the direction and speed of a toboggan. For example, if a toboggan is being pulled at an angle to the ground, the force of the pull will have a vertical component that helps to pull the toboggan downhill, as well as a horizontal component that helps to steer the toboggan in a particular direction.

3. How does the weight of a person on a toboggan affect the ride?

The weight of a person on a toboggan can affect the ride in several ways. A heavier person will increase the overall weight of the toboggan, which can affect the speed and momentum of the ride. Additionally, a heavier person may also experience more friction and air resistance, which can impact the overall speed and distance traveled.

4. What type of surface is best for tobogganing?

The best surface for tobogganing is a smooth, packed snow surface. This type of surface provides a good balance of friction and speed, allowing the toboggan to move smoothly and quickly down the hill. Ice or wet surfaces can be too slippery, while deep or powdery snow can slow down the toboggan too much.

5. How can you increase the speed and distance of a toboggan ride?

To increase the speed and distance of a toboggan ride, you can reduce the friction and air resistance by using a smooth, packed snow surface. You can also decrease the weight of the toboggan by removing excess weight or using a lighter toboggan. Additionally, having a steep slope and a long, straight track can also help to increase the speed and distance of the ride.

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