{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Applications - Applications Sample Problem 5-7 Figure 5.6...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Applications Sample Problem 5-7 Figure 5.6 shows a block of mass m = 15 kg hanging from three cords. What are the tensions in these cords ? The mass m experiences a gravitational force equal to mg. Since the mass is at rest, cord C must provide an opposing force equal to mg. Applying Newton's third law, we conclude that cord C exerts a force on the knot whose magnitude is equal to mg (and pointed in the direction shown in Figure 5.6). Since the system is at rest, the net force on the knot must be equal to zero: This vector equation can be rewritten in terms of its components along the x-axis and y-axis, using the following information: Figure 5.6. Sample Problem 5-7. Using these expressions we can write down the equations for the x and y-components of the net force: The first expression can be used to express T A in terms of T B : Substituting this expression into the equation for [Sigma]F y we obtain: from which we can calculate T B : Knowing T B , we can now calculate T A :
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
In the case of sample problem 5-6, the tensions in the cords are: T A = 100 N T B = 140 N T C = 150 N Problem Figure 5.7 shows a block with mass m on a frictionless plane, tilted by an angle [theta]. What is the acceleration of the block ? Figure 5.7. Mass m on an inclined plane. In order to determine the acceleration of the block we have to determine the total force acting on the block along the inclined plane. Two forces act on the block: the gravitational force exerted by the earth on the block, and a force, called the normal force exerted by the plane on the block (see Figure 5.8). This force must be present since in its absence mass m will experience free fall
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern