A block of mass m rests on a smooth plane of a wedge which is inclined at angle of θ to the horizontal. The wedge is accelerating to the left such that the block does not slide down. Derive an expression for the normal reaction force on the block and the acceleration of the wedge.
Ans given: normal reaction force = mg / cos θ and acceleration = g tan θ.
As I read your question, the wedge is moving such that it cancels the downward graviith force on the block.
Downward force on block due to gravity=mg*sinTheta
upward force along the plane due to wedge moving=ma*cosTheta
set them equal
ma*cosTheta=mg*sinTheta
a=g*tanTheta
Downward force: I don't know which force you want here.
To derive the expressions for the normal reaction force and the acceleration of the wedge, we can analyze the forces acting on the block and the wedge separately.
Let's consider the block first. The forces acting on the block are its weight mg and the normal reaction force N. The weight acts vertically downward and can be resolved into two components: one parallel to the inclined plane (mg*sinθ) and one perpendicular to the inclined plane (mg*cosθ).
Since the block is not sliding down, the friction force between the block and the wedge must be equal to the component of the weight parallel to the plane (mg*sinθ).
The normal reaction force (N) acting on the block is equal to the component of the weight perpendicular to the inclined plane (mg*cosθ).
Now, let's consider the forces acting on the wedge. Since the system is accelerating to the left, there must be a net force acting to the left. The horizontal component of the normal reaction force N is responsible for this net force.
Now, using Newton's second law, we can write the equations of motion for the block and the wedge.
For the block:
Consider the forces along the inclined plane (x-axis direction),
Sum of forces = mass of the block * acceleration of the block
mg*sinθ - friction force = m*a (equation 1)
For the wedge:
Consider the forces along the horizontal direction,
Sum of forces = mass of the wedge * acceleration of the wedge
Net force due to horizontal component of N = M*A (equation 2)
Now, let's solve these equations.
Since the block is on a smooth plane of the wedge, there is no friction force (friction = 0), so equation 1 simplifies to:
mg*sinθ = m*a
Solving for a, we get:
a = g*sinθ
Now, let's focus on equation 2. The normal reaction force N can be calculated by considering the vertical forces on the wedge:
Sum of vertical forces = 0
N - mg*cosθ = 0
Solving for N, we get:
N = mg*cosθ
Therefore, the normal reaction force on the block is N = mg*cosθ, as given in the answer.
Substituting the value of N into equation 2, we get:
Net force due to horizontal component of mg*cosθ = M*A
Simplifying further, we can write:
mg*cosθ = M*A
Dividing both sides by M, we get:
A = g*cosθ
But we need to express the acceleration in terms of the given angle θ, so we use the trigonometric identity:
tanθ = sinθ / cosθ
Rearranging the equation, we get:
cosθ = sinθ / tanθ
Substituting this value into the equation for A, we get:
A = g*tanθ
Therefore, the acceleration of the wedge is A = g*tanθ, as given in the answer.