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 Î˛ < 90Â°.

#### 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.

##### In this diagram, we look at force F_1 only.

##### In this diagram, we look at force F_2

### With the help of the diagrams, we can expressÂ 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 thatÂ \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