Answer:
33.4
Explanation:
Step 1:
\sumMo=0 (moment about the origin)
Fb(15)-Fc(15)=0
Fb=Fc
Step 2:
\sumFx=0
-Fb-Fccos\theta+Ncsin\theta=0
Fc=0.3Nc=Fb
-0.3Nc-0.3Nccos\theta+Ncsin\theta=0
(-0.3-cos\theta+sin\theta)Nc=0----(1)
\sumFy=0
Nccos\theta+Fcsin\theta-Nb=0
Nccos\theta+0.3Ncsin\theta-Nc=0
Nc[cos\theta+0.3sin\theta-1]=0--------(2)
Solving eq (1) and eq (2)
\theta=33.4
Step 3:
As the roller is a two force member
2(90-\phi)+\theta=180
\phi=\theta/2
\phi=Tan(\muN/N)-1
\phi=16.7
\theta=2x16.7=33.4
The largest angle θ of the incline so that the roller does not slip for any force P applied to the beam is; θ = 33.4°
Taking anti - clockwise moment about the center of the roller, we have;
ΣM₀ = 0; (F_b × 15) - (F_c × 15) = 0 -----(eq 1)
ΣFₓ = 0; -F_b - (F_c × cos θ) + (N_c × sin θ) = 0 ----(eq 2)
ΣF_y = 0; (N_c × cos θ) + (F_c × sin θ) - N_b = 0 ------(eq 3)
We are given coefficient of static friction as; μ_b = μ_c = 0.3 and if we assume that the roller slips at point C, then; F_c = 0.3N_c
From eq(1) we can simplify to get;
F_b = F_c
Thus;
F_b = 0.3N_c
And our eq(2) becomes;
-0.3N_c - (0.3N_c × cos θ) + (N_c × sin θ) = 0
⇒ N_c(-0.3 - 0.3cos θ + sin θ) = 0 ----(eq 4)
Dividing both sides by N_c gives;
(-0.3 - 0.3cos θ + sin θ) = 0
Solving for θ gives; θ = 33.4°
Making N_b the subject in eq(3) gives;
(N_c × cos θ) + (F_c × sin θ) = N_b
Since F_c = 0.3N_c, then;
(N_c × cos θ) + (0.3N_c × sin θ) = N_b
⇒ N_c(cos θ + 0.3sin θ) = N_b
⇒ N_c(cos 33.4° + 0.3sin 33.4°) = N_b
Since eq. 4 is satisfied for any value of N_c, then any value of P can act on the beam. We also know that the roller is a two-force member and so the largest possible angle would still be calculated to be θ = 33.4°.
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