Can Two Street Cars Be Pulled Up a Slope with a Low Friction Coefficient?

In summary, the cars are going up a slope and the first car has fallen off a bridge. The second car is going up the slope and has on its breaks. People rush to jump on the second car so it will be heavy enough to pull up the first car. The initial acceleration of the cars is ft/s^2.
  • #1
cheez
26
0
Two identical street cars are connected by a cable. One car has fallen off a bridge. The other car is going up a slope of 14 degrees, and has on its breaks, mu=0.13. People rush to jump on the second car so it will be heavy enough to pull up the first car. What is the initial acceleration of the cars, in ft/s^2
After drawing the graph, I still dun know how to solve the problem. I know I should use F=ma, but I really dun know about tension. btw, am I drawing the right sketch? thanks so much!
car.JPG
 
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  • #2
That's correct
goto http://knowhowstuff.blogspot.com/search/label/Physics%20Q%20and%20A
for more details
 
Last edited:
  • #3
I think the car is supposed to be going uphill; your car is going downhill..right?
 
  • #4
nop it going up
thats why
F=m*a=friction+ tension+w*g*sin(theta) is upwards
 
  • #5
Which direction is friction acting in?
I think you have it right in your picture, but you use it incorrectly when you're resolving forces in your equation.
 
  • #6
since motion is upwards, the friction acts on downwards.
 
  • #7
uskalu said:
since motion is upwards, the friction acts on downwards.

I don't know who came up withthis problem or its solution, but it appears seriously flawed. Even if the 2nd car was on a level track, the brakes couldn't prevent it from sliding off the bridge along with the first car, if the coefficient of static friction was only 0.13. And then even assuming this is an incorrect coefficient, then in order to pull both cars up the slope, there would have to be a friction force up the slope, applied between the train's driving wheels and the tracks, to move it at some constant speed, not at some minimum acceleration, and the max force available would depend upon then mass of the car and the passengers in it. Bad problem.
 

1. What is a tension graph problem?

A tension graph problem is a physics problem that involves analyzing the forces acting on an object or system and determining the tension in each of its supporting strings or cables. These types of problems are often encountered in mechanics and engineering courses.

2. How do you draw a tension graph?

To draw a tension graph, first identify all the forces acting on the object or system and draw them as arrows pointing in the direction of the force. Then, label each force with its corresponding tension value. Finally, draw a line representing the string or cable and label it with the tension value at each end. The graph should show the direction and magnitude of each force and the tension in the supporting strings or cables.

3. What is the purpose of a tension graph?

The purpose of a tension graph is to visually represent the forces acting on an object or system and the tensions in its supporting strings or cables. This allows for a better understanding of the dynamics of the system and can help in solving problems involving tension and forces.

4. How do you solve a tension graph problem?

To solve a tension graph problem, first draw the graph and label all the forces and tensions. Then, apply Newton's laws of motion to set up equations and solve for the unknown tensions. It is important to consider the direction and magnitude of each force and to use the appropriate equations for the given scenario.

5. What are some common mistakes when solving tension graph problems?

Some common mistakes when solving tension graph problems include forgetting to consider the direction of forces, using incorrect equations, and not labeling the graph properly. It is important to carefully analyze the problem and the given information before attempting to solve it. Additionally, double-checking the solution and ensuring that it makes sense in the context of the problem can help avoid mistakes.

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