Q.278. Statics

Question 278.
A massless rod has a mass glued to it. The position of the mass is 1 m from the left end and 3 m from the right end. This rod-mass combo is placed on the frictionless inclined planes as shown. At equilibrium, it makes an angle 'B' with the horizontal. What is this angle?

Answer 278.
Refer to the figure as under.

Let P be the lower end of the rod where the normal reaction is N and Q be the higher end of the rod where the normal reaction is N'

Three forces act on the rod,
(i) weight Mg of the block attached to it,
(ii) normal reaction N at P and
(iii) normal reaction N' at Q.

Balancing horizontal components of these three forces,
Ncos30° = N'cos60° ... (1)
Taking moments of the forces about P,
Mg * 1cosB = N' * 4cos(30° - B) ... (2)

Taking moments of forces about Q,
Mg * 3cosB = N * 4sin(30° - B) ... (3)

From (2) and (3),
3N' = N tan(30° - B) ... (4)

From (1) and (4),
3 N' = N' [cos60° / cos30°] tan(30° - B)
=> tan(30° - B) = 3 * [(√3/2) / (1/2)] = 3√3
=> 30° - B = arctan(3√3) = 79.11°
=> B = - 49.11°.

ALTERNATE SOLUTION:

As the angle B works out negative, it means that the inclination of the rod is reverse of what is shown in the question. The revised solution leading to the same answer with reversed inclination of the rod alongwith the figure is as under. Refer to the figure shown below.

Let P be the upper left end of the rod where the normal reaction is Nand Q be the lower right end of the rod where the normal reaction is N'.

Let PQ be at an angle B above the horizontal at Q.

Three forces act on the rod,

(i) weight Mg of the block attached to it,

(ii) normal reaction N at P and

(iii) normal reaction N' at Q.

Balancing horizontal components of these three forces,

Ncos30° = N'cos60° ... (1)

Taking moments of the forces about P,

Mg * 1cosB = N' * 4sin(60° - B) ... (2)

Taking moments of forces about Q,

Mg * 3cosB = N * 4cos(60° - B) ... (3)

From (2) and (3),

tan(60° - B) = (1/3) (N/N') = (1/3) (cos60°/cos30°) ... [from (1)]

=> tan(60° - B) = 1/(3√3)

=> 60° - B = arctan (1/3√3) = 10.89°

=> B = 49.11°.

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