I have a question that goes thus. At what angle $\theta$ must the $500N$ force be applied in order that the resultant $R = 1000N$. For this condition what will.ne the angle $\beta$ between $R$ and horizontal

I got

$ \theta =97.9^\circ$ and $\beta=34.1^\circ$

But I'm not sure that's rightly done. I need help with verification of result

my attemptenter image description here Here's my solution

  • $\begingroup$ This looks like a homework exercise. Please show what you've tried so far. $\endgroup$ – NMech Mar 7 at 4:22
  • 1
    $\begingroup$ You can verify your answer using graphic method. $\endgroup$ – r13 Mar 7 at 4:25

The problem in your solution is a confusion about angle $\theta$ from the cosine rule you are using. The triangle of forces will look like the following image:

enter image description here

where r is the angle in the cosine rule:

$$R^2 = F_1^2 +F_2^2 - 2F_1 F_2 \cos(r)$$

$$R^2 = F_1^2 +F_2^2 - 2F_1 F_2 \cos(180-\theta)$$

If this is superimposed to your initial image it will look like the following:

enter image description here

So what you calculate as 97.9 is actually equal to $180-\theta$. Therefore $\theta$ in your original drawing is $82.096[deg]$.

When you substitute, you can obtain the components of $R_x, R_y$ (notice the use of $\phi$). Therefore:

  • $R_x = F_1 + F_2 \cos(\phi) = -868.75[N]$
  • $R_y = F_2 \sin(\phi) = 495.25[N]$ (Negative because its pointing to the left)

If you take the magnitude of $R$ you get $\sqrt{(-868.75)^2 + 495.25^2}=1000$ as specified.

Regarding the angle between R (lets denote it $\phi_R$) and the positive X axis:

$$\tan(\phi_R)=\frac{R_y}{R_x} = \frac{495}{-868.75} $$ $$\phi_R=150[deg]$$

enter image description here

Another way

I used the following method to find the angle $\theta$. It leads to the same values.

Lets write $F_1 = -800[N]$ and $F_2=500N$. Also assume the $\phi$ is angle with the positive x, therefore $\phi = 180-\theta$.

Then the resultant R will have:

$$R_x = F_1 + F_2 \cos(\phi) \qquad R_y=F_2 \sin(\phi)$$

Then $$|R|= \sqrt{(F_1 + F_2 \cos(\phi))^2 + (F_2 \sin(\phi))^2 }$$

Solving for this you get

$$\phi= 97.9[deg]\rightarrow \theta = 82.1$$

  • 1
    $\begingroup$ If we draw a x-y grid system, and set the tail end of the forces at the origin, both forces are falling in the fourth quadrant, as well as the resultant force. The angle 97.09 degrees is correct, if measured CCW from the positive x-axis towards the 500 N force. The angle between the two forces, therefore, is 180 - 97.09 = 82.91 degrees.. $\endgroup$ – r13 Mar 7 at 6:17
  • $\begingroup$ @Opeolluwa I managed today to read your solution, and reply to the question with respect to where things went wrong in your calculation. $\endgroup$ – NMech Mar 8 at 13:34

Since we are working on force vectors. You can should always verify your answers graphically with direction in mind. The diagram below is an example utilizing the original sketch and using parallelogram method. Note that in which the vectors (500N & 800N) shall be drawn to correct length and direction, thus the resultant R can be directly measured off

enter image description here

In addition, the angle between the resultant force and the horizontal x-axis can be solved by triangular method (solving the right triangle).

[![enter image description here][2]][2]

Please let me know, if the solves above contain mistakes. [2]: https://i.stack.imgur.com/lMfVD.png


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