# Preventing a block from sliding down a frictionless inclined plane

• yesmale4
In summary: You don’t learn much from a summary of the correct answer.You asked for a summary of the conversation, which I provided. If you want an explanation of why you got the wrong answer, you should provide your working and specific questions for clarification. Please keep in mind that this forum is for helping students learn, not for providing complete solutions or doing their work for them.
yesmale4
Homework Statement
A block of mass 1.8 kg is placed on a frictionless inclined ramp of mass 12.8 kg, with the angle
θ= 34 as shown below.
Calculate the size of the net horizontal force that must be applied to the ramp to prevent the block from accelerating with respect to the ramp?
Relevant Equations
F=ma
hey i would like to understand something, i solve this question but i don't understand why my answer is right, first of all we learn that in problems like this we need to disassemble into components the mg and that what i try to do here but i didnt get the right answer so then i try to do same but this time on the N (normal) and i was correct i add my answer and i hope for some explanations

It's not clear what you don't understand. Personally, I would use a reference frame accelerating to the right with acceleration ##a##. In this frame, we have fictitious forces of ##-ma## and ##-Ma## on both masses, along with gravity and the normal forces. These forces must balance on both masses for relative equilibrium between the blocks.

Last edited:
Orodruin

You can do this problem relatively simply by noting that the total of gravitational force and the fictitious force on the block must be perpendicular to the ramp (otherwise they cannot exactly cancel the normal force from the ramp and you won’t have equilibrium). There will be only one acceleration ##a## for which this is the case. The force on the block will be whatever force is necessary to give the full system this acceleration.

PeroK

Your analysis is based on the fact that since the normal force ##N## is the only force acting on the block of mass ##m_1## in the horizontal direction, it is the net horizontal force on it and is therefore responsible for its acceleration ##a##. Here is a streamlined version of how to put it together in the inertial frame of the surface on which the motion takes place.

1. In vertical direction, the block of mass m1 has zero acceleration so its weight must be balanced by the vertical component of the normal force. This gives ##N\cos\!\gamma=m_1g.##
2. In the horizontal direction the acceleration is ##a##, therefore ##N \sin\!\gamma=m_1a##.
3. Finally, since the two masses accelerate as one and the net force on the system of two masses is ##F##, it follows that ##F=(m_1+m_2)a##.

This gives a system of 3 equations and 3 unknowns, ##N##, ##a## and ##F##, which you have solved to find ##F##. On your way there, you also found ##N## and ##a## in terms of the given quantities.

A word of caution: If you have more than one masses, use subscripts, else you might get hopelessly confused if the problem is more involved than this.

On edit: Modified the text to clarify it according to @Orodruin's suggestion in #5.

Last edited:
PeroK
kuruman said:
Your analysis is based on the fact that since the normal force N is the only force acting on the block of mass m1,
... in the horizontal direction. Important distinction.
kuruman said:
A word of caution: If you have more than one masses, use subscripts, else you might get hopelessly confused if the problem is more involved than this.
It seems to me that the notation is given by the problem formulation. OP can hardly be faulted for this.

kuruman
yesmale4 said:
... but i didnt get the right answer so then i try ...
If you want to know why you originally got the wrong answer, you need to post the working which gave the wrong answer.

Identifying and understanding mistakes is an important part of learning.

Orodruin

## 1. What is the purpose of preventing a block from sliding down a frictionless inclined plane?

The purpose of preventing a block from sliding down a frictionless inclined plane is to understand the concept of friction and its role in preventing motion. It also helps in studying the principles of Newton's Laws of Motion and their application in real-life scenarios.

## 2. How does friction affect the motion of a block on an inclined plane?

Friction is the force that opposes the motion of an object. In the case of a block on an inclined plane, friction acts in the opposite direction of the block's motion, making it more difficult for the block to slide down the plane. This results in the block remaining stationary or moving at a slower rate.

## 3. What factors can affect the amount of friction on an inclined plane?

The amount of friction on an inclined plane can be affected by several factors, such as the weight of the block, the angle of the inclined plane, and the surface materials of the block and the inclined plane. The roughness of the surfaces and the presence of any lubricants can also impact the amount of friction.

## 4. How can we prevent a block from sliding down a frictionless inclined plane?

To prevent a block from sliding down a frictionless inclined plane, we can increase the angle of the plane, add weight to the block, or use a rougher surface for either the block or the plane. We can also use external forces, such as applying a force in the opposite direction of the block's motion, to counteract the effects of friction.

## 5. What are the real-life applications of studying the motion of a block on an inclined plane?

The study of a block's motion on an inclined plane has several real-life applications. It helps in understanding the mechanics of objects sliding down ramps, hills, or other inclined surfaces. This knowledge is also crucial in designing and building structures, such as roads and bridges, that can withstand the effects of friction and gravity. Additionally, it has applications in fields such as engineering, physics, and sports, where friction plays a significant role in the motion of objects.

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