Question

Question: Using strings and pulleys, three weights of mas 6kg, 8kg and 10kg are suspended in equilibrium as shown. Calculate the magnitude of the angle ACB.

The textbook's answer is 90 degrees.

I have tried through two approach: 1) making vertical components of tensions forces equal to the weight force 2) rearranging tension forces and the weight force to calculate the angle using cosine rule. 1st approach gives me 88.83 degrees, while the second computes the answer to be 90 degrees.

Answer 1

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Since the system is in equilibrium;

The moment of forces about point A should be zero

Thus;

10g×0.5+15g×1=Tsin53° ×1

50+150=0.8T

T=250 N

T=250 NAt point A, the hinge force is working which can be resolved as Ry& Rx

Rx=Tcos 53°

Rx =250×0.6=150 N

Ry =Tsin53°

Ry =250×0.8=200 N

R=(Rx )2 +(Ry)^2

R=150^2 +200^2

R=250 N

Therefore, The tension in the string will be 250 N and the hinged force at point A will be 250 N

Answer 2

$\huge \pink{ \boxed{ \green{ \mathfrak{ \fcolorbox{red}{aqua}{ \underline{ \red{ ꪖꪀᦓ᭙ꫀ᥅}}}}}}}$

Since the system is in equilibrium;

The moment of forces about point A should be zero

Thus;

10g×0.5+15g×1=Tsin53° ×1

50+150=0.8T

T=250 N

T=250 NAt point A, the hinge force is working which can be resolved as Ry& Rx

Rx=Tcos 53°

Rx =250×0.6=150 N

Ry =Tsin53°

Ry =250×0.8=200 N

R=(Rx )2 +(Ry)^2

R=150^2 +200^2

R=250 N

Therefore, The tension in the string will be 250 N and the hinged force at point A will be 250 N