Coefficients of friction:
The coefficients of friction under the sliding supports of the door in Figure are equal to 0.30 at A and 0.20 at B. What forces P shall give the 25 kg door a left ward acceleration of 3 m/sec^{2}.
Solution
The force, that means actions and reactions acting on the door are marked in Free Body Diagram. After showing the position and direction of inertia force on the body, we may write equilibrium equations as below :
∑ F_{X}= 0, + P - X _{A} - X _{B }- 25 a = 0
∑ F_{Y}= 0, + Y_{A} + Y_{B} - 25 g = 0
∑ M_{ CG} = 0
Y_{A} × 1.2 + X_{ A} × 1.8 - Y_{B} × 1.2 + X _{B} × 1.8 + P × 0.6 = 0
Also we have
X _{A} = 0.3 Y_{A}
X _{B} = 0.2 Y_{B}
And a = 3 m / sec ^{2}
∴ P = 0.3 Y_{A} + 0.2 Y_{B} + 75 ------------ (1)
and, 1.74 Y_{A} - 0.84 Y_{B} + 0.6 P = 0 ------- (2)
∴ 0.6 P = - 1.74 Y_{A} + 0.84 Y_{B}
∴ P =- 2.90 Y_{A} + 1.4 Y_{B} ---------- (3)
Comparing Eqs. (1) and (3), we get
P = 0.3 Y_{A} + 0.2 Y_{B} + 75 = - 2.90 Y_{A} + 1.4 Y_{B}
∴ 3.2 Y_{A} - 1.2 Y_{B} = - 75 --------- (4)
Also we have (from Σ F_{y} = 0) :
Y_{A} + Y_{B } = 25 × g
= 245.25 -------- (5)
After solving Eqs. (4) and (5), we obtain
Y_{B} = 195.41 N
∴ Y_{A} = 49.84 N
And P = 129.04 N