Question

In the figure u is large enough, so that the block does not slip on the wedge. The contact force between the wedge and the block is (g = 10 m/s2) 2kg O = 30° Ancuar​

Answer 1

Answer:

Let the acceleration of system is a

P=(M+m)a

=12a

F.B.D of block w.r.t wedge is

Balancing the force

masin53=mgsin37

N=mgcos37+macos53

a=10∗

5

3

∗

4

5

=7.5m/s

2

F=12∗7.5

=90N

Explanation:

this helps you

Answer 2

Given:

Angle of inclination θ = 30°

Mass of block = 2 kg

Gravitational acceleration = 10 $m/s^{2}$

To Find: Contact force between block and wedge.

Calculation:

The figure shown below has the following forces:

• A = Normal force perpendicular to the surface of contact.
• B = Vector component of 'mg' = mg cos θ
• D = Vector component of 'mg' = mg sin θ
• E = force due to gravity = mg
• Block does not slip, which means C and D are equal in magnitude.
• Contact force = A = Normal Force = B = mg cos θ = 2*10* cos 30 = 10√3 Newton

Answer:

The contact force between wedge and the block is 10√3 Newtons.