# Acceleration acting on a block lying on a wedge (non-inertial frame)

#### Like Tony Stark

Homework Statement
Consider this situation: there's a wedge, where a block is lying on it. There's no friction. A horizontal acceleration is applied to the wedge. This acceleration may cause three cases: the block doesn't move with respect to the wedge; the block slides up; the block slides down.
Homework Equations
Newton's equation
I have some difficulties trying to understand non-inertial frames.

I have problems to notice the acceleration in these cases, from an inertial reference frame and from non inertial refrence frame.

Consider the first case, if I'm on the wedge, I see that the block doesn't move so there's no acceleration, all the forces add up to zero. But what about if I'm on an inertial frame? Which acceleration would the block have? Just horizontal acceleration, right? (Because I would see that it is moving to the right, so if I consider non inclined axis, the acceleration would be just in $x$)

And what about the case where it is sliding up? From the non inertial frame, I would see just $x$ acceleration (if I consider inclined axis). And what if I saw it from the ground? What components would it have?

Related Introductory Physics Homework Help News on Phys.org

#### kuruman

Homework Helper
Gold Member
Say the contact is frictionless. If the wedge slides to the right with acceleration $\vec a$ in the inertial frame, an observer in the non-inertial frame will see the block experience a fictitious force $-m\vec a$. If the block is at rest relative to the wedge in the non-inertial frame, the observer will conclude that the fictitious force is just enough to keep the block from sliding down. This situation is equivalent to the one in which the block is not accelerating in an inertial problem, but is prevented from sliding down by a real force, e.g. a finger, holding it in place. Sliding up or sliding down means that the fictitious (or real) force is greater or less than what is needed to keep the block from sliding.

#### Like Tony Stark

Say the contact is frictionless. If the wedge slides to the right with acceleration $\vec a$ in the inertial frame, an observer in the non-inertial frame will see the block experience a fictitious force $-m\vec a$. If the block is at rest relative to the wedge in the non-inertial frame, the observer will conclude that the fictitious force is just enough to keep the block from sliding down. This situation is equivalent to the one in which the block is not accelerating in an inertial problem, but is prevented from sliding down by a real force, e.g. a finger, holding it in place. Sliding up or sliding down means that the fictitious (or real) force is greater or less than what is needed to keep the block from sliding.
Yes, I understand that. But I don't understand the acceleration of the block in the second case. From an inertial frame, it will just have $x$ acceleration? Or $x$ and $y$? And from non inertial? Will it have just $x$ acceleration?

#### haruspex

Homework Helper
Gold Member
2018 Award
Yes, I understand that. But I don't understand the acceleration of the block in the second case. From an inertial frame, it will just have $x$ acceleration? Or $x$ and $y$? And from non inertial? Will it have just $x$ acceleration?
It will have the acceleration you found in the wedge frame of reference plus the acceleration of the wedge.

"Acceleration acting on a block lying on a wedge (non-inertial frame)"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving