Solving a 100kg Trolley Uphill Problem: Can You Help?

  • Thread starter Morker
  • Start date
In summary, to move the trolley up the incline at a constant velocity, the force F must be equal to the force of friction (110N) plus the component of the weight parallel to the incline (mg*sin(30) = 100kg*9.8m/s^2*sin(30) = 490N). Thus, F = 110N + 490N = 600N. To accelerate the trolley up the incline at a rate of 2m/s^2, the net force must be equal to ma, so F = ma = 100kg*2m/s^2 = 200N. To calculate the acceleration when F = 1000N, we can rearrange
  • #1
Morker
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Homework Statement


!00kg trolley is being pushed up a rough 30 degree incline by a constant force F the firction force Ff between the incline and the trolley is 110n gravity is 9.8ms

Homework Equations


Determine the value of F that will move the trolley up the incline at a constant velocity
determine the value of F that will accelerate the trolley up the incline at a value of 2ms
calculate the acceleration if F=1000
what is the value of F if they trolley accelerates up the incline at 10ms

The Attempt at a Solution


Given this question at school have no formulas on how to do completely lost have spent 2 hours trying to find info on the net can't find anything
i attempted first question with mgxsin30* + 110 and got 600N and that was correct answer but i don't think i have the correct working out
and 2nd question i just did fnet = 100x2 600+200 = 800N and that was correct answer but I am pretty sure working out is wrong
 
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  • #2
You have the correct answers, but please show how you arrived at them. For part a, since the movement is at constant velocity, there must be no net force acting in the direction of the incline (Newton 1, sum of all forces parallel to incline = 0). For part b, the net force acting in the direction of the incline must be equal to 'ma' (Newton 2, sum of all forces parallel to incline = ma).
 

Related to Solving a 100kg Trolley Uphill Problem: Can You Help?

1. How do you determine the force needed to push a 100kg trolley uphill?

The force needed to push a trolley uphill depends on several factors including the weight of the trolley, the angle of the incline, and the coefficient of friction between the trolley and the surface. This can be calculated using the formula F = mg(sinθ + μcosθ), where F is the force needed, m is the mass of the trolley, g is the acceleration due to gravity, θ is the angle of the incline, and μ is the coefficient of friction.

2. Is it possible to push a 100kg trolley uphill without using external force?

Yes, it is possible to push a trolley uphill without using external force if the incline is steep enough. This is because the force of gravity will act upon the trolley and pull it downwards, causing it to move uphill.

3. How does the angle of the incline affect the force needed to push a 100kg trolley uphill?

The steeper the incline, the greater the force needed to push the trolley uphill. As the angle of the incline increases, the force of gravity acting on the trolley also increases, making it more difficult to push the trolley upwards.

4. Can the coefficient of friction be ignored when solving a 100kg trolley uphill problem?

No, the coefficient of friction cannot be ignored as it plays a crucial role in determining the force needed to push the trolley uphill. The higher the coefficient of friction, the more force will be needed to overcome the resistance and move the trolley uphill.

5. Are there any other factors that should be considered when solving a 100kg trolley uphill problem?

Aside from the weight of the trolley, the angle of the incline, and the coefficient of friction, other factors that should be considered include the surface of the incline (e.g. roughness, slope), the condition of the trolley (e.g. wheels, brakes), and the strength and stamina of the person pushing the trolley.

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