Q.363. Static equilibrium

Question 363.


Determine the horizontal and vertical   
components of reaction at the pin A
and the normal force at the smooth
peg B on the member.

Answer 363.
 
Let R = normal reaction at peg B.

Taking moments of forces about A,
R * (0.40) = 600 * (0.80) cos30° ... [angle between AC and normal to the direction of F = 30°]
=> R = 1039 N.

Let Rx and Ry be the horizontal (to the right) and vertical (upwards) components of reaction at pin A.

Balancing horizontal components of all the forces,
Rx
= Rcos60° + Fcos30°
= 1039 * cos60° - 600 * cos30°
= - 0.115 N
[Negative sign indicates that the horizontal reaction at A is to the left.]

Balancing vertical components of all the forces,
Ry
= Fsin30° - Rsin60°
= 600 * sin30° - 1039 * sin60°
= - 599.8 N
[Negative sign indicates that the vertical reaction at the pin A is downwards.]

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