# Rope pulling a block problem

• kaotic
In summary, the conversation is about a problem involving a 5.00 kg block being pulled along a frictionless floor by a cord with a force of 12.0 N at an angle of 25 degrees above horizontal. The problem asks for the magnitude of the block's velocity and the value of the force just before the block is lifted off the floor. The conversation also includes someone's progress on the problem and a request for help. The solution involves finding the value of the force when the normal force is zero and using the vertical component of the tension to balance out the weight of the block. The block's acceleration can be found using the same method as part a.

#### kaotic

This is the first time I've posted here. I am having a problem with an assigned problem. I know the policy here is to not give out answers, and I'm not looking for that. I just need a starting point. Here's the problem:

A 5.00 kg block is pulled along a horizontal frictionless floor by a cord that exerts a force of magnitude 12.0 N at an angle of 25 degrees above horizontal. a) make a free-body diagram and determine the magnitude of the block's velocity b) the F magnitude is slowly increased. What is the value (of the force, I'm assuming) just before the block is lifted (completely) off the floor? c) What is the magnitude of the block's acceleration just before it is lifted (completely) off the floor.

OK, I have done part a), drawing the free body diagram and figured out the magnitude of the acceleration (+2.17m/s^2). I also know that to do b), I need to find the value of the force when the normal force is equal to zero. From there, I can't figure out where to go.

If someone could help me out with an equation to use and/or other tips, it would be much appreciated.

Chris

I'm assuming that you don't have to worry about the block pivoting about its connection point to the string for this problem, so:

You know that the normal force will be zero in part b. The only other forces acting on the block are its weight and the tension of the string. This means that the vertical component of the tension will exactly balance out the weight. Since you know one component of the tension and the angle the string is at, you can use that to find F = Fy/sin(25).

The block's acceleration is then found the same way as part a: there is 0 acceleration in the y direction and the x-component of the tension provides an acceleration in that direction.

Hi Chris,

Welcome to the forum! It's great that you're looking for help with your assigned problem. As you mentioned, the policy here is to not give out answers, but we can definitely provide you with some guidance and tips to help you solve the problem.

First, let's review the problem and make sure we understand it correctly. We have a 5.00 kg block being pulled along a horizontal frictionless floor by a cord with a force of 12.0 N at an angle of 25 degrees above horizontal. We need to find the magnitude of the block's velocity (part a), the value of the force just before the block is lifted off the floor (part b), and the magnitude of the block's acceleration just before it is lifted off the floor (part c).

For part a, you have already correctly determined the magnitude of the block's acceleration to be +2.17 m/s^2. To find the magnitude of the block's velocity, you can use the equation v = u + at, where v is the final velocity, u is the initial velocity (which we can assume is 0 since the block starts from rest), a is the acceleration, and t is the time. In this case, we know the acceleration and we can find the time using the equation d = ut + 1/2at^2, where d is the distance, u is the initial velocity (again, 0), a is the acceleration, and t is the time. We know the distance in this case, which is the length of the cord being pulled, and we can solve for t. Once we have t, we can plug it into the first equation to find the final velocity.

For part b, you're correct that you need to find the value of the force when the normal force is equal to zero. This is because the normal force is what keeps the block in contact with the floor, so when it reaches zero, the block will be lifted off the floor. To find this value, you can use the equation F = ma, where F is the force, m is the mass (which we know is 5.00 kg), and a is the acceleration (which we know from part a). Plug in the values and solve for F.

For part c, we can use the same equation as part b, F = ma, but this time we need to find the acceleration when the normal force is equal to zero.

## 1. How does the number of people pulling the rope affect the force required to move the block?

The more people pulling the rope, the less force each individual needs to exert to move the block. This is because the total force applied is divided among the number of people pulling.

## 2. What is the relationship between the angle of the rope and the force required to move the block?

The force required to move the block is directly proportional to the angle of the rope. As the angle increases, the force required also increases.

## 3. What role does friction play in the rope pulling a block problem?

Friction between the block and the surface it is resting on can make it more difficult to move the block, as it creates resistance against the pulling force. Additionally, the coefficient of friction between the block and the surface can affect the amount of force required to overcome this resistance.

## 4. How does the weight of the block impact the force needed to move it?

The weight of the block affects the force needed to move it. The heavier the block, the more force is required to overcome its inertia and move it. This is why it may be easier for more people to pull a heavier block.

## 5. Is the force required to move the block the same in all directions?

No, the force required to move the block may vary depending on the direction in which the rope is being pulled. For example, if the rope is pulled at an angle, the force required will be greater than if the rope is pulled in a straight line parallel to the surface the block is resting on.