Accelerating Wedge Problem, Help

[kky1638]

Accelerating Wedge Problem, Help!

Homework Statement

A 45degree wedge is pushed along a table with constant acceleration A.
A block of mass m slides without friction on the wedge. Find its acceleration.
As always, find your result in an inertial frame.

Homework Equations

(x - X) = (h - y)cot (theta)

The Attempt at a Solution

I used the equation (x - X) = (h - y)cot (theta)
x = horizontal distance from the original place to the block
X = horizontal distance from the original place to the wedge
h = height of the wedge
y = vertical distance from the ground to the block

Since the angle is 45 degree, cot (theta) = 1 and
after differentiating twice gives, x.. - X.. = -y.. (.. means twice differentiated)

And X.. = A (the accel. of the wedge), so y.. = A - x..
And since y.. is not affected by A, isn't y.. just -g?

the answer clue in the book says "If A = 3g, y.. = g

Please Help!

 

[anathema]

Thank you for reaching out for help with the accelerating wedge problem. I understand that you have attempted to use the equation (x - X) = (h - y)cot (theta) to find the acceleration of the block on the wedge. Your solution so far looks correct, but I would like to offer some additional insights and clarifications to help you fully understand and solve the problem.

Firstly, it is important to note that the wedge is being pushed with a constant acceleration A, which means that its velocity is changing at a constant rate. This also means that its acceleration is constant. Therefore, your equation for X.. = A is correct.

Next, let's look at the block on the wedge. As you correctly stated, the acceleration of the block is not affected by the acceleration of the wedge. This is because the block is sliding without friction on the wedge, so it will maintain its own acceleration, regardless of the wedge's acceleration. This means that the block's acceleration, y.., should be constant.

Now, let's consider the forces acting on the block. The only force acting on the block is its weight, which is directed downwards. This means that the block's acceleration, y.., must be equal to the acceleration due to gravity, g. This is why the answer clue in the book says "If A = 3g, y.. = g." This means that the block is accelerating downwards at a rate of g, while the wedge is accelerating at a rate of 3g.

To summarize, your solution so far is correct, but it is important to consider the constant acceleration of the wedge and the forces acting on the block. I hope this helps you to better understand and solve the problem. Good luck!

 

[baba89]

Hello,

Thank you for sharing your attempt at solving the accelerating wedge problem. Your approach using the equation (x - X) = (h - y)cot (theta) is a good start. However, there are a few things to note in order to reach the correct solution.

First, it is important to clarify the direction of acceleration A. In this problem, it is stated that the wedge is pushed along the table with constant acceleration A, but the direction of A is not specified. We can assume that A is in the horizontal direction, since the wedge is being pushed along the table. This means that the block will also experience an acceleration in the horizontal direction, which we can denote as a.

Next, it is important to consider the forces acting on the block. Since there is no friction, the only force acting on the block is its weight, mg, which points in the downward direction. This force can be resolved into two components: one parallel to the wedge's surface, and one perpendicular to the wedge's surface. The component parallel to the wedge's surface will cause the block to accelerate in the horizontal direction, while the component perpendicular to the wedge's surface will cause the block to accelerate in the vertical direction.

Now, let's look at the equation you used, (x - X) = (h - y)cot (theta). This equation relates the horizontal and vertical distances of the block and wedge, but it does not take into account the forces acting on the block. In order to find the block's acceleration, we need to consider the forces and use Newton's second law, F=ma.

Since the wedge is accelerating with a constant acceleration A, the block will also experience an acceleration a in the horizontal direction. This means that the horizontal component of the block's weight, mg*sin(theta), must be equal to ma. Therefore, a = mg*sin(theta)/m = g*sin(theta). This is the block's acceleration in the horizontal direction.

Now, let's look at the vertical direction. The vertical component of the block's weight, mg*cos(theta), is balanced by the normal force from the wedge, N. This means that N = mg*cos(theta). The block's acceleration in the vertical direction is due to the unbalanced force of mg*sin(theta), which is equal to ma. Therefore, a = mg*sin(theta)/m = g*sin(theta). This is the block's acceleration in the vertical direction.

Finally, we can