If the resultant force acting on the bracket is directed


If the resultant force acting on the bracket is directed along the positive y axis, determine the magnitude of the resultant force and the coordinate direction angles of F so that β < 90°.

If the resultant force acting on the bracket is directed

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

Solution:

One way to do this question would be to express each of the forces as x,y,z components. To do that, a diagram would be highly beneficial. Thus, let us draw out the vector components as follows. We drew large diagrams so that you can clearly see what is happening in each.

If the resultant force acting on the bracket is directed solution

In this diagram, we look at force F_1 only.

If the resultant force acting on the bracket is directed solution

In this diagram, we look at force F_2

 

With the help of the diagrams, we can express forces F and F_1 in a Cartesian vector form.

F_1=600\text{cos}30^0\,\text{sin}30^0(+i)+600\text{cos}30^0\,\text{cos}30^0(+j)+600\text{sin}30^0(-k)

F_1=\left\{259.81i+450j-300k\right\}N

F_2=500\text{cos}\alpha i+500\text{cos}\beta j+500\text{cos}\gamma k

 

Also note that the resultant force is going to be directed along the positive y axis as stated by the question, that means that as a Cartesian vector component, we can write:

F_R=F_R\,j

 

To find the resultant force, we only need to add up the corresponding vector forms.

F_R=F_1+F_2

F_R\,j=(259.81i+450j-300k)+(500\text{cos}\alpha i+500\text{cos}\beta j+500\text{cos}\gamma k)

F_R\,j=(259.81+500\text{cos}\alpha)i+(450+500\text{cos}\beta)j+(-300+500\text{cos}\gamma)k

 

Now, we will solve for \alpha , \beta , \gamma by equating the i, j, and k components.

0=259.81+500\text{cos}\alpha

\alpha=121.31^0

 

0=-300+500\text{cos}\gamma

\gamma=53.13^0

 

F_R=450+500\text{cos}\beta

 

Remember that \text{cos}^2\alpha +\text{cos}^2\beta +\text{cos}^2\gamma =1

 

Using this equation, we can substitute \alpha and the \gamma value to figure out F_R and \beta

\text{cos}\beta =\pm \sqrt{1-\text{cos}^{2}(121.31^0)-\text{cos}^{2}(53.13^0)}

\text{cos}\beta =\pm 0.6083

 

Now that we know what \beta is, we can substitute it back into our equation.

F_R=450+500(0.6083)=754N

\beta=cos^{-1}(0.6083)=52.5^0

 

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-71.

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