A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is :- [2018]A: B: C: a = g cos θD: a = g tan θ
Question
A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is :- [2018]A: B: C: a = g cos θD: a = g tan θ
Solution
To solve this problem, we need to consider the forces acting on the block.
- The weight of the block (mg) acts vertically downwards.
- The normal force (N) from the wedge acts perpendicular to the surface of the wedge.
- The friction force (f) acts along the surface of the wedge, opposing the motion of the block.
Since the block is not moving, the net force acting on it must be zero. This means that the component of the weight acting along the wedge must be balanced by the friction force, and the component of the weight acting perpendicular to the wedge must be balanced by the normal force.
Let's break down the weight into its components:
- The component of the weight acting along the wedge is mg sin θ.
- The component of the weight acting perpendicular to the wedge is mg cos θ.
Since the block is not moving, the friction force must be equal to the component of the weight acting along the wedge, and the normal force must be equal to the component of the weight acting perpendicular to the wedge.
Now, let's consider the motion of the wedge. Since it is accelerating to the right with acceleration 'a', there must be a net force acting on the block in the opposite direction (to the left) due to the inertia of the block. This force is equal to ma.
For the block to remain stationary on the wedge, this inertial force must be balanced by the component of the weight acting along the wedge. Therefore, we have:
ma = mg sin θ
Solving this equation for 'a' gives:
a = g sin θ
So, the correct answer is not listed in the options provided. The relation between 'a' and 'θ' for the block to remain stationary on the wedge is a = g sin θ.
Similar Questions
A block of mass m is placed on a smooth wedge of mass M placed at rest on a smooth horizontal surface as shown in the figure. During the time the block slides on the wedge,The centre of mass of the “block-wedge” system accelerates downward by acceleration a < g.Work done by the contact forces between block and wedge is zero in any frame.Momentum of the block-wedge system remains zero in horizontal direction.Work done by the contact forces between block and wedge is zero only in ground frame.
A block of mass m is placed on a smooth wedge of mass M. The value of force F such that m does not slide down relative to wedge is
A block of mass M slides down a frictionless plane inclined at an angle θ with the horizontal. The gravitational force on the block is directedGroup of answer choicestoward the center of the Earth.toward the center of mass of the block.perpendicular to the plane.parallel to the plane in the same direction as the movement of the block.parallel to the plane in the opposite direction as the movement of the block
Consider the situation in the diagram. Block 1 has mass m1and is on a slope at an angle θ from horizontal. Block 2 hasmass m2 .Find an expression for the acceleration of the blocks, ignor-ing friction.You will be marked on the following:1) Free body diagram for block 1.2) Free-body diagram for block 2.3) Relevant equations for block 1.4) Relevant equations for block 2.5) Solving the equations for acceleration.
A small block slides with velocity 0.5√gr on the horizontal frictionless surface as shown in the Figure. The block leaves the surface at point C. The angle θ in the Figure is:
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.