What is the weight of the sled being pulled up a ramp by a girl?

  • Thread starter Thread starter Statik
  • Start date Start date
  • Tags Tags
    Box Ramp
Click For Summary
SUMMARY

The problem involves a girl with a mass of 50 kg pulling a sled up an 8° slope with a coefficient of static friction of 0.155. The maximum force of friction calculated is 75.21 N, while the gravitational force acting down the slope is 68.19 N, resulting in a net force of 7.02 N. The girl can achieve an acceleration of 0.030 m/s², leading to an incorrect sled weight calculation of 1681 kg. The error lies in not properly analyzing the net forces acting on the sled, particularly the tension in the rope and the sled's weight component down the slope.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of static friction and its calculation
  • Ability to draw and interpret free body diagrams
  • Familiarity with trigonometric functions in physics
NEXT STEPS
  • Review the principles of free body diagrams in physics
  • Learn how to apply Newton's second law (F_net = ma) in various scenarios
  • Study the effects of tension in ropes on inclined planes
  • Explore the relationship between friction and weight on slopes
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and forces, as well as educators seeking to clarify concepts related to motion on inclined planes.

Statik
Messages
1
Reaction score
0

Homework Statement



A girl of mass mg = 50 kg is pulling a sled up a slippery slope. The coefficient of friction between the girl's boots and the slope is µs = 0.155; the friction between the sled and the slope is negligible. The girl can pull the sled up the slope with a ≤ amax = 0.030 m/s2 before she begins to slip. Assume the rope connecting the girl to the sled is kept parallel to the slope at all times. The angle of the slope is θ = 8°.

Homework Equations





The Attempt at a Solution



I have already worked out the maximum value of the force of friction on the girl as:
50kg *9.8 m/s^2 = 490N
Mg(j-hat) = COS 8 * 490N = 485.23N
Fs = Us * Mg(jhat) = 0.155 * 485.23N = 75.21N
Gravity Force on Girl, down the ramp = Sin 8 * 490N = 68.19N

Net force unused by girl is75.21N - 68.19N = 7.02N
Net acceleration is 0.03m/s^2 so, using remaining force, 7.02N / 0.03m/s^2 = 234kg
Total box should be 234kg / sin 8 = 1681 kg but that isn't right. I know my total friction for the girl and I know it's right . . . what am I missing in my calculation of the box weight?

Thanks!
Rod
 
Physics news on Phys.org
Statik said:

Homework Statement



A girl of mass mg = 50 kg is pulling a sled up a slippery slope. The coefficient of friction between the girl's boots and the slope is µs = 0.155; the friction between the sled and the slope is negligible. The girl can pull the sled up the slope with a ≤ amax = 0.030 m/s2 before she begins to slip. Assume the rope connecting the girl to the sled is kept parallel to the slope at all times. The angle of the slope is θ = 8°.

Homework Equations


The Attempt at a Solution



I have already worked out the maximum value of the force of friction on the girl as:
50kg *9.8 m/s^2 = 490N
Mg(j-hat) = COS 8 * 490N = 485.23N
Fs = Us * Mg(jhat) = 0.155 * 485.23N = 75.21N
Gravity Force on Girl, down the ramp = Sin 8 * 490N = 68.19N

Net force unused by girl is75.21N - 68.19N = 7.02N
Net acceleration is 0.03m/s^2 so, using remaining force, 7.02N / 0.03m/s^2 = 234kg
Total box should be 234kg / sin 8 = 1681 kg but that isn't right. I know my total friction for the girl and I know it's right . . . what am I missing in my calculation of the box weight?

Thanks!
Rod
You did well I in determining the tension in the rope of 7 N, but then you did not analyze the net force acting on the sled correctly. The tension acts up the slope on the sled, but the component of its unknown weight acts down the slope. Try your F_net = ma equation again with a good free body diagram. Like you did when analyzing the forces on the girl.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
5
Views
17K
Replies
2
Views
2K
Replies
5
Views
12K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K