Resolving vector equations to find angle











up vote
1
down vote

favorite












Say we have a beam hinged to a wall. The beam is length $l$, and of mass $2m_1$. The other end of the beam is supported by a rope which goes over a pulley and is connected to a weight of mass $m_2$. We have three forces $vec{H}$ being the force on the hinge, $vec{M}$ being the force of weight of the beam and $vec{R}$ being the tension of the rope. Below is a diagram showing the situation on the left, and the forces on the right.



Situation diagram on left, forces on the right



We take the origin $o$ to be from the hinge on the wall. We specify the axis with positive $x$ right along the beam and $y$ up from this. The rope forms an angle $theta$ with the beam. This system is in equilibrium. We're now asked to find $theta$ in terms of the parameters given. I have resolved the vectors as follows:



$$vec{H}=-m_1gcot{theta}(vec{x})+m_1g(vec{y})$$
$$vec{R}=m_1gcot{theta}(vec{x})+m_1g(vec{y})$$
$$vec{M}=-2m_1g(vec{y})$$



How do I resolve this into an equation for $theta$?










share|cite|improve this question


























    up vote
    1
    down vote

    favorite












    Say we have a beam hinged to a wall. The beam is length $l$, and of mass $2m_1$. The other end of the beam is supported by a rope which goes over a pulley and is connected to a weight of mass $m_2$. We have three forces $vec{H}$ being the force on the hinge, $vec{M}$ being the force of weight of the beam and $vec{R}$ being the tension of the rope. Below is a diagram showing the situation on the left, and the forces on the right.



    Situation diagram on left, forces on the right



    We take the origin $o$ to be from the hinge on the wall. We specify the axis with positive $x$ right along the beam and $y$ up from this. The rope forms an angle $theta$ with the beam. This system is in equilibrium. We're now asked to find $theta$ in terms of the parameters given. I have resolved the vectors as follows:



    $$vec{H}=-m_1gcot{theta}(vec{x})+m_1g(vec{y})$$
    $$vec{R}=m_1gcot{theta}(vec{x})+m_1g(vec{y})$$
    $$vec{M}=-2m_1g(vec{y})$$



    How do I resolve this into an equation for $theta$?










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Say we have a beam hinged to a wall. The beam is length $l$, and of mass $2m_1$. The other end of the beam is supported by a rope which goes over a pulley and is connected to a weight of mass $m_2$. We have three forces $vec{H}$ being the force on the hinge, $vec{M}$ being the force of weight of the beam and $vec{R}$ being the tension of the rope. Below is a diagram showing the situation on the left, and the forces on the right.



      Situation diagram on left, forces on the right



      We take the origin $o$ to be from the hinge on the wall. We specify the axis with positive $x$ right along the beam and $y$ up from this. The rope forms an angle $theta$ with the beam. This system is in equilibrium. We're now asked to find $theta$ in terms of the parameters given. I have resolved the vectors as follows:



      $$vec{H}=-m_1gcot{theta}(vec{x})+m_1g(vec{y})$$
      $$vec{R}=m_1gcot{theta}(vec{x})+m_1g(vec{y})$$
      $$vec{M}=-2m_1g(vec{y})$$



      How do I resolve this into an equation for $theta$?










      share|cite|improve this question













      Say we have a beam hinged to a wall. The beam is length $l$, and of mass $2m_1$. The other end of the beam is supported by a rope which goes over a pulley and is connected to a weight of mass $m_2$. We have three forces $vec{H}$ being the force on the hinge, $vec{M}$ being the force of weight of the beam and $vec{R}$ being the tension of the rope. Below is a diagram showing the situation on the left, and the forces on the right.



      Situation diagram on left, forces on the right



      We take the origin $o$ to be from the hinge on the wall. We specify the axis with positive $x$ right along the beam and $y$ up from this. The rope forms an angle $theta$ with the beam. This system is in equilibrium. We're now asked to find $theta$ in terms of the parameters given. I have resolved the vectors as follows:



      $$vec{H}=-m_1gcot{theta}(vec{x})+m_1g(vec{y})$$
      $$vec{R}=m_1gcot{theta}(vec{x})+m_1g(vec{y})$$
      $$vec{M}=-2m_1g(vec{y})$$



      How do I resolve this into an equation for $theta$?







      trigonometry physics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 21 at 16:13









      whitelined

      586




      586






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          You've set up your equations wrong. Let's start with the weight of the beam. I think you call it $vec M_1$ in your figure, but $vec M$ in the equations. So let's stick with the equation. That's correct. Now, since the beam is in equilibrium, the force on the hinge, the weight of the beam, and the force in the rope add to $0$:
          $$vec H+vec M+vec R=0$$
          Once you calculate $vec R$, you can use this equation to calculate the force on the hinge. But that's not what they ask.



          I have no idea where the equation for $vec R$ is coming from. Since it is a tension in an inextensible string, it means the magnitude is constant in the string. The other end, where you suspend mass $m_2$, the string is in equilibrium, so $|vec R|=m_2 g$



          We need now one more equation. For that, note that the beam is not rotating around the hinge, so the total torque is zero. Note that only forces perpendicular to the beam have torques. So it will be $vec M$ in the negative $y$ direction and the $y$ component of $vec R$
          $$-|vec M|frac l2+|vec R|sintheta l=0$$
          Putting it all together, $g$ and $l$ cancel, so you have $$-frac{2m_1}2+m_2sintheta=0$$ or $$sintheta=frac{m_1}{m_2}$$






          share|cite|improve this answer





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007938%2fresolving-vector-equations-to-find-angle%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            You've set up your equations wrong. Let's start with the weight of the beam. I think you call it $vec M_1$ in your figure, but $vec M$ in the equations. So let's stick with the equation. That's correct. Now, since the beam is in equilibrium, the force on the hinge, the weight of the beam, and the force in the rope add to $0$:
            $$vec H+vec M+vec R=0$$
            Once you calculate $vec R$, you can use this equation to calculate the force on the hinge. But that's not what they ask.



            I have no idea where the equation for $vec R$ is coming from. Since it is a tension in an inextensible string, it means the magnitude is constant in the string. The other end, where you suspend mass $m_2$, the string is in equilibrium, so $|vec R|=m_2 g$



            We need now one more equation. For that, note that the beam is not rotating around the hinge, so the total torque is zero. Note that only forces perpendicular to the beam have torques. So it will be $vec M$ in the negative $y$ direction and the $y$ component of $vec R$
            $$-|vec M|frac l2+|vec R|sintheta l=0$$
            Putting it all together, $g$ and $l$ cancel, so you have $$-frac{2m_1}2+m_2sintheta=0$$ or $$sintheta=frac{m_1}{m_2}$$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              You've set up your equations wrong. Let's start with the weight of the beam. I think you call it $vec M_1$ in your figure, but $vec M$ in the equations. So let's stick with the equation. That's correct. Now, since the beam is in equilibrium, the force on the hinge, the weight of the beam, and the force in the rope add to $0$:
              $$vec H+vec M+vec R=0$$
              Once you calculate $vec R$, you can use this equation to calculate the force on the hinge. But that's not what they ask.



              I have no idea where the equation for $vec R$ is coming from. Since it is a tension in an inextensible string, it means the magnitude is constant in the string. The other end, where you suspend mass $m_2$, the string is in equilibrium, so $|vec R|=m_2 g$



              We need now one more equation. For that, note that the beam is not rotating around the hinge, so the total torque is zero. Note that only forces perpendicular to the beam have torques. So it will be $vec M$ in the negative $y$ direction and the $y$ component of $vec R$
              $$-|vec M|frac l2+|vec R|sintheta l=0$$
              Putting it all together, $g$ and $l$ cancel, so you have $$-frac{2m_1}2+m_2sintheta=0$$ or $$sintheta=frac{m_1}{m_2}$$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                You've set up your equations wrong. Let's start with the weight of the beam. I think you call it $vec M_1$ in your figure, but $vec M$ in the equations. So let's stick with the equation. That's correct. Now, since the beam is in equilibrium, the force on the hinge, the weight of the beam, and the force in the rope add to $0$:
                $$vec H+vec M+vec R=0$$
                Once you calculate $vec R$, you can use this equation to calculate the force on the hinge. But that's not what they ask.



                I have no idea where the equation for $vec R$ is coming from. Since it is a tension in an inextensible string, it means the magnitude is constant in the string. The other end, where you suspend mass $m_2$, the string is in equilibrium, so $|vec R|=m_2 g$



                We need now one more equation. For that, note that the beam is not rotating around the hinge, so the total torque is zero. Note that only forces perpendicular to the beam have torques. So it will be $vec M$ in the negative $y$ direction and the $y$ component of $vec R$
                $$-|vec M|frac l2+|vec R|sintheta l=0$$
                Putting it all together, $g$ and $l$ cancel, so you have $$-frac{2m_1}2+m_2sintheta=0$$ or $$sintheta=frac{m_1}{m_2}$$






                share|cite|improve this answer












                You've set up your equations wrong. Let's start with the weight of the beam. I think you call it $vec M_1$ in your figure, but $vec M$ in the equations. So let's stick with the equation. That's correct. Now, since the beam is in equilibrium, the force on the hinge, the weight of the beam, and the force in the rope add to $0$:
                $$vec H+vec M+vec R=0$$
                Once you calculate $vec R$, you can use this equation to calculate the force on the hinge. But that's not what they ask.



                I have no idea where the equation for $vec R$ is coming from. Since it is a tension in an inextensible string, it means the magnitude is constant in the string. The other end, where you suspend mass $m_2$, the string is in equilibrium, so $|vec R|=m_2 g$



                We need now one more equation. For that, note that the beam is not rotating around the hinge, so the total torque is zero. Note that only forces perpendicular to the beam have torques. So it will be $vec M$ in the negative $y$ direction and the $y$ component of $vec R$
                $$-|vec M|frac l2+|vec R|sintheta l=0$$
                Putting it all together, $g$ and $l$ cancel, so you have $$-frac{2m_1}2+m_2sintheta=0$$ or $$sintheta=frac{m_1}{m_2}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 21 at 17:04









                Andrei

                10.4k21025




                10.4k21025






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007938%2fresolving-vector-equations-to-find-angle%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Bundesstraße 106

                    Verónica Boquete

                    Ida-Boy-Ed-Garten