Question
Asked Nov 6, 2019

Two blocks A and B, each having a mass of 3.5 kg , are connected by the linkage shown. Neglect the weight of the links. (Figure 1) 

If the coefficient of static friction at the contacting surfaces is μs = 0.5, determine the largest force P that can be applied to pin C of the linkage without causing the blocks to move.

30°
В
30°
А
30°
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30° В 30° А 30°

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Expert Answer

Step 1

Consider Block A. Applying the equilibrium conditions, we get:

ΣΕ-0
FI cos 300uN= 0
F cos 30°0.5N, = 0
AC
N 2F Cos30°
A
Σ-0
F sin 300 N -W 0
F sin 30° 2Fc cos30° mg 0
FC (sin 30° 2 cos 30°)-(3.5x9.81) 0
34.335
F
AC
sin 30° 2 cos 30°
F
= -27.868 N
AC
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ΣΕ-0 FI cos 300uN= 0 F cos 30°0.5N, = 0 AC N 2F Cos30° A Σ-0 F sin 300 N -W 0 F sin 30° 2Fc cos30° mg 0 FC (sin 30° 2 cos 30°)-(3.5x9.81) 0 34.335 F AC sin 30° 2 cos 30° F = -27.868 N AC

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Step 2

Consider Block B. Applying the equilibrium conditions, we get:

ΣΗ-0
-WN3 sin 60° - uNsin30° = 0
В
-(3.5x 9.81)N2(sin60° - 0.5sin 30°) = 0
34.335
В
sin 60° 0.5 sin 30°
NB 55.736 N
В
Σ-0
F. =
NR cos 60°
Fr + uN3 cos 30° 0
55.736 cos 60° Fc 0.5(55.736) cos30° 0
BC
F 52.003 N
BC
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ΣΗ-0 -WN3 sin 60° - uNsin30° = 0 В -(3.5x 9.81)N2(sin60° - 0.5sin 30°) = 0 34.335 В sin 60° 0.5 sin 30° NB 55.736 N В Σ-0 F. = NR cos 60° Fr + uN3 cos 30° 0 55.736 cos 60° Fc 0.5(55.736) cos30° 0 BC F 52.003 N BC

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Step 3

Consider Joint C. Applying the equi...

ΣΕ-0
Psin 300 F F sin60° 0
Psin 300-52.003 - (-27.868) sin 60 ο = 0
BC
AC
P= 55.737 Ν
ΣΕ-0
Pcos 300 F
cos 60° = 0
Pcos 300 27.868) cos 60° = 0
P= 16.09 Ν
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ΣΕ-0 Psin 300 F F sin60° 0 Psin 300-52.003 - (-27.868) sin 60 ο = 0 BC AC P= 55.737 Ν ΣΕ-0 Pcos 300 F cos 60° = 0 Pcos 300 27.868) cos 60° = 0 P= 16.09 Ν

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