Pulley Problem, 1 crate hanging, 1 crate on an incline

In summary, the conversation discusses a physics problem involving two blocks of different masses connected by a cord over a frictionless pulley. The question asks for the acceleration of the hanging block and the tension in the cord. A free body diagram is drawn and equations involving tension and acceleration are set up. To solve the problem, the equations must be solved simultaneously for both variables. The conversation also mentions the need for notes on this topic.
  • #1
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Here is a question that I am not sure where to start.
A block of mass m1 = 6.07 kg on a frictionless plane inclined at angle = 33.3° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.46 kg hanging vertically (Fig. 5-54). (a) What is the acceleration of the hanging block (choose the positive direction up)? (b) What is the tension in the cord?

So far, I have drawn a free body diagram. I will call the crate on the incline m1, and the crate that is hanging m2. So to start, I have drawn a free body diagram. For m2, there is the weight of the crate (M x g), and the tension in the cord that is suspending the crate. Therefore, the tension should be equal to the weight of the box, correct? Next for m1, I altered the axes to corrsepond with the incline. The weight is still acting straight downwards, while the force that is sliding the crate down the incline is equal to (Weight x sin 33.3 degrees). There is also in the tension acting towards the edge of the incline which is opposite of the force acting down the incline. (?). So I have all of these forces, so to find the acceleration, do I simply add the weight of m1 and m2, and also add the tension of wire from the crate that is dangling, then use the mass of m2 to find the accleration? And for the second part of the question, I just unsure what to do, if i just add up the tension on the wire from crates m1 and m2? Thank you for the help
 
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  • #2
Tension is less than weight of crate. You have 2 equations involving T and a:
Mg – T = Ma (for crate) &
T – mgsin(33.3°) = ma (mass on slope).
Now all you have to do is solve these 2 simultaneous equations for T & a. Don't you have any notes on this?
 
  • #3


I would begin by clarifying the given information and assumptions. It is stated that the plane is frictionless, but it is not specified whether the pulley is also frictionless. This could affect the calculations for the tension in the cord.

Next, I would use Newton's second law of motion, F=ma, to analyze the forces acting on each crate. For m1, the forces are the weight (mg) acting down the incline and the tension in the cord acting parallel to the incline. The net force in the x-direction (parallel to the incline) can be found by using the component of the weight that is parallel to the incline, which is mg sinθ. This force is equal and opposite to the tension in the cord. So, the net force in the x-direction is 0, and the acceleration in that direction is also 0.

For m2, the forces are the weight (mg) acting downwards and the tension in the cord acting upwards. The net force in the y-direction (perpendicular to the incline) can be found by using the component of the weight that is perpendicular to the incline, which is mg cosθ. This force is equal to the tension in the cord. So, the net force in the y-direction is mg cosθ, and the acceleration in that direction is mg cosθ/m2.

To find the total acceleration, we can use the Pythagorean theorem to find the magnitude of the acceleration vector. So, the total acceleration is √(0^2 + (mg cosθ/m2)^2) = mg cosθ/m2.

To find the tension in the cord, we can use the net force equation again, but this time for m2. So, the tension in the cord is equal to the weight of m2, which is mg.

In summary, the acceleration of the hanging block is mg cosθ/m2 and the tension in the cord is mg. It is important to note that these calculations are based on the assumption of a frictionless pulley. If the pulley does have some friction, it would need to be taken into account in the calculations. Additionally, it would be helpful to double check the calculations and make sure all units are consistent.
 

1. How do pulleys work in a pulley problem?

Pulleys work by using a combination of fixed and movable wheels to change the direction and magnitude of force applied to an object. In a pulley problem, the pulley system is used to lift or move objects with less effort.

2. How does the weight of the crates affect the pulley problem?

The weight of the crates affects the pulley problem by determining the amount of force needed to lift or move them. The heavier the crates, the more force is required to overcome their weight and move them using the pulley system.

3. What is the relationship between the incline and the force needed to move the crate?

The steeper the incline, the more force is needed to move the crate. This is because the weight of the crate is acting in the direction of the incline, making it harder to overcome. As the incline becomes less steep, less force is needed to move the crate.

4. How does friction affect the pulley problem?

Friction can affect the pulley problem by adding resistance to the movement of the crates. If there is high friction between the crates and the surface they are on, more force will be needed to move them. Reducing friction, for example by using lubricants, can make it easier to move the crates with the pulley system.

5. Is there an optimal number of pulleys to use in a pulley problem?

The optimal number of pulleys to use in a pulley problem depends on various factors such as the weight of the objects, the angle of the incline, and the amount of friction. Generally, using more pulleys can reduce the force needed to move the object, but it also increases the complexity of the system. It is important to carefully consider these factors to determine the most efficient number of pulleys to use.

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