If the magnitude of the resultant force acting on the bracket is to be 450 N directed along the positive u axis, determine the magnitude of F1 and its direction Φ.

#### Solution:

### We will first draw the vector components acting on the bracket as follows.

### The dashed arrows represent the x and y components of each force. Force, F_2, (in blue) is along the x axis, thus it has no y component.

### Now, we will draw the resultant force (as stated in the question) and it’s components also.

### We can now write down the x and y component values of each force.

### (F_1)_x=F_1\text{ sin}\phi

### (F_1)_y=F_1\text{ cos}\phi

### (F_2)_x=200\,N

### (F_2)_y=0

### (F_3)_x=260(\frac{5}{13})=100\,N

### (F_3)_y=260(\frac{12}{13})=240\,N

### (F_R)_x=450\text{ cos 30}^0=389.71\,N

### (F_R)_y=450\text{ sin 30}^0=225\,N

### The next step is to sum the forces along the x and y axes. To do this, we will the establish the positive sides. We will pick forces acting up, and forces acting to the right to be positive.

### +\rightarrow\sum(F_R)_x=\sum(F_x)

### 389.71=F_1\text{ sin}\phi +200+100

### F_1\text{ sin}\phi =89.71———————————(1)

### +\uparrow\sum(F_R)_y=\sum(F_y)

### 225=F_1\text{ cos}\phi -240

### F_1\text{ cos}\phi =465———————————(2)

### All that is left is to solve equations (1) and (2) simultaneously. To do so, remember the identity, \frac{\text{sin}\theta}{\text{cos}\theta}=\text{tan}\theta

why is it 389.71=f1 cos thi +200 + 100

the x component of f3 does not equal 100

There was a typo, which has been fixed. Many thanks for that. F3 does indeed equal 100. Please carefully look over the solution again. Hope it helps 🙂