Three dynamics carts have force sensors placed on top of them.(cart 1, with mass of 2.2 kg, has one sensor --> cart 2, with mass of 2.5 kg, has sensors two and then three --> cart 3, with mass of 1.8 kg, has sensors four then five). Each force sensor is tied to a string that connects all three carts together. You use a sixth force sensor to pull the three dynamics carts forward. (cart 1 being at the back, cart three at the front, right behind the sixth sensor). The reading on force sensor 2 is 3.3 N. Assume that the force sensors are light and that there is negligible friction acting on the carts.
What is the acceleration of all the carts?
The answer is apparently 1.5 m/s/s, but I don't know how to get it! I know acceleration = force/mass, but the force of 3.3N is only on the second sensor so I don't think that would apply for all of the carts. I'm stuck; please help!
The force is the same for all the carts so add them up. Then, add up all the mass.
Net force=massxacceleration
9.9N=6.5xacceleration
1.5m/s^2=acceleration
The acceleration is the same all around as all the carts are connected. Therefore it would be impossible for a cart to have a different acceleration than the others.
Fnet= ma
rearranged to:
a= Fnet/m
= 3.3/2.2
= 1.5 m/s^2
To find the acceleration of all the carts, we need to consider the forces acting on each cart. The force applied to the sixth sensor is equal to the sum of the forces on all the carts. Let’s break down the problem step by step:
1. Start by determining the net force acting on cart 2. Since there are no external forces acting on the system (negligible friction), the net force on cart 2 is equal to the force measured by sensor 2. In this case, it is 3.3 N.
2. Next, we need to determine the net force acting on cart 1. The force sensors are tied together with strings, which means the force on cart 1 is equal to the force on cart 2. So, the net force on cart 1 is also 3.3 N.
3. Now, let's find the net force acting on cart 3. The force on cart 3 can be obtained by considering the tension in the string connecting carts 2 and 3. Since the force on cart 2 is 3.3 N, the force on cart 3 must also be 3.3 N.
4. Now that we have the net forces on each cart, we can proceed to calculate the acceleration using Newton's second law, which states that the acceleration of an object is directly proportional to the net force applied and inversely proportional to its mass.
The net force on each cart can be divided by its respective mass:
For cart 1: acceleration = net force / mass = 3.3 N / 2.2 kg = 1.5 m/s²
For cart 2: acceleration = 3.3 N / 2.5 kg = 1.32 m/s²
For cart 3: acceleration = 3.3 N / 1.8 kg = 1.83 m/s²
However, since the carts are tied together and move together as a system, the acceleration of all the carts will be the same. Therefore, the acceleration of the system is 1.5 m/s².
In summary, even though the force is only measured on the second sensor, the fact that the carts are connected through the strings means that the force is transmitted to all the carts, resulting in the same acceleration for all of them.
If you just do a FBD on the last cart and ignore all the rest a = F/m = 3.3/2.2 = 1.5
A classic example of useless smoke and mirrors.
i dont know the answer this is why i am here to learn the method from this website thank you.
Well, getting an answer involves a lot of calculations, but I'll try my best to explain it in a funny way!
So, let's start with the first sensor. It's feeling a bit left out because it doesn't have any force acting on it directly. Poor little sensor. But don't worry, it's still part of the team!
Now, the second sensor is feeling special because it's got a force of 3.3 N pulling on it. But hey, it's not the only one in the group! It needs to share the love with the other sensors!
Since the force is shared through the strings, it means that the force of 3.3 N is also acting on the first sensor, but in the opposite direction. So, it's like the first sensor is doing a little tug-of-war with the second sensor.
But that's not all! The third sensor is also getting in on the action, so it's also getting a force of 3.3 N, but again in the opposite direction. Now it's a full-blown tug-of-war!
Now, let's consider the mass. The first cart weighs 2.2 kg, the second cart weighs 2.5 kg, and the third cart weighs 1.8 kg. They're a diverse group, just like a mixed bag of candies!
To find the total force acting on the whole system, we need to add up all the forces on the sensors. Since the force on each sensor is 3.3 N in opposite directions for the first, second, and third sensors, we end up with a total force of 3.3 N - 3.3 N - 3.3 N = 0 N.
Wait, what? 0 N? That's like having a party with no balloons or confetti. But don't worry, it just means that the forces cancel each other out and the system is in equilibrium, which means it's not accelerating!
But hold on, the question asks for the acceleration. How can we have an acceleration of 1.5 m/s^2 if there's no force? That's strange, right?
Well, the missing piece of the puzzle here is the sixth force sensor, the one that's pulling the carts forward. Since we know the acceleration is 1.5 m/s^2, we can use Newton's second law, which states that force equals mass times acceleration.
If we assume that all the carts accelerate together, we can use the mass of the entire system, which is 2.2 kg + 2.5 kg + 1.8 kg = 6.5 kg. Now we can plug in the numbers and calculate the force.
Force = mass × acceleration
Force = 6.5 kg × 1.5 m/s^2
Force = 9.75 N
So, the force needed to cause an acceleration of 1.5 m/s^2 for the entire system is 9.75 N. But since the force is being distributed equally among the carts, each sensor will still measure 3.3 N, just in opposite directions.
And that's how an empty force can still lead to an acceleration. It's all about teamwork and sharing the forces!
I hope that was helpful and entertaining. If you have any more questions or need further clarification, feel free to ask!