Question

Mechames of Machines Two identical rollers, each of weight Q = 100N are supported by an inclined plane and vertical wall. Assuming smooth surfaces, find the reaction inclined at the point of support A, B and C​

Answer 1

Step-by-step explanation:

Since the rollers are identical and the surfaces are smooth, the reaction forces at points A and C must be equal and vertical.

Let's call this force R. At point B, the reaction force is perpendicular to the inclined plane, so we can use trigonometry to find its magnitude.

Let's call this force T. Using the forces in the vertical direction, we can write: 2R + T = 2Q Using the forces in the horizontal direction

we can write: T = R tanθ where θ is the angle of inclination of the plane. Substituting the second equation into the first, we get: 2R + R tanθ = 2Q Simplifying: R (2 + tanθ) = 2Q R = 2Q / (2 + tanθ)

Plugging in the values: R = 2(100N) / (2 + tanθ) We can use trigonometry to find the value of tanθ: tanθ = opposite / adjacent = BC / AB

Since the rollers are identical, BC = AB, so: tanθ = AB / AB = 1 Therefore, R = 2(100N) / (2 + 1) = 50N and T = R tanθ = 50N * 1 = 50N

So the reaction forces at points A and C are both equal to 50N and vertical, while the reaction force at point B is equal to 50N and perpendicular to the inclined plane.