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down the ramp? Fgx unit n Pick a direction from below no direction, the magnitude is zero no direction, it is scalar

down the ramp? Fgx unit n Pick a direction from below no direction, the magnitude is zero no direction, it is scalar

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Part B**: What is the component of the force of gravity pulling \(m_2\) down the ramp? \[ F_{gx} = \, \underline{\hspace{3em}} \, \text{unit} \, n \] - **Pick a direction from below:** - Green arrow: \(\downarrow\) - Purple arrow: \(\leftarrow\) - Red arrow: \(\rightarrow\) - Red arrow pointing down-right - Red arrow pointing up-right - No direction, the magnitude is zero - No direction, it is scalar **Diagram**: A series of arrows indicate possible directions for the gravitational component. Each arrow is associated with a radio button for selection. --- **Part C**: What must the minimum force of static friction be to keep \(m_2\) from moving? \[ F_{f} = \, \underline{\hspace{3em}} \, \text{unit} \, n \] - **Pick a direction from below:** - Green arrow: \(\downarrow\) - Purple arrow: \(\leftarrow\) - Red arrow: \(\rightarrow\) - Red arrow pointing down-right - Red arrow pointing up-right - No direction, the magnitude is zero - No direction, it is scalar **Diagram**: Similar to Part B, arrows indicate possible directions for the static friction force with a checkbox for selection next to the upward right arrow. --- **Part D**: What is the minimum coefficient of static friction? (Text for Part D is present, but not further expanded in the image.)
Transcribed Image Text:**Part B**: What is the component of the force of gravity pulling \(m_2\) down the ramp? \[ F_{gx} = \, \underline{\hspace{3em}} \, \text{unit} \, n \] - **Pick a direction from below:** - Green arrow: \(\downarrow\) - Purple arrow: \(\leftarrow\) - Red arrow: \(\rightarrow\) - Red arrow pointing down-right - Red arrow pointing up-right - No direction, the magnitude is zero - No direction, it is scalar **Diagram**: A series of arrows indicate possible directions for the gravitational component. Each arrow is associated with a radio button for selection. --- **Part C**: What must the minimum force of static friction be to keep \(m_2\) from moving? \[ F_{f} = \, \underline{\hspace{3em}} \, \text{unit} \, n \] - **Pick a direction from below:** - Green arrow: \(\downarrow\) - Purple arrow: \(\leftarrow\) - Red arrow: \(\rightarrow\) - Red arrow pointing down-right - Red arrow pointing up-right - No direction, the magnitude is zero - No direction, it is scalar **Diagram**: Similar to Part B, arrows indicate possible directions for the static friction force with a checkbox for selection next to the upward right arrow. --- **Part D**: What is the minimum coefficient of static friction? (Text for Part D is present, but not further expanded in the image.)
The two masses shown below are connected by a cable and are **not moving**. Answer the following questions about the system. The diagram shows two masses connected by a cable over a pulley. - The first mass \( m_1 \) is labeled as 25.1 kg and is hanging vertically. - The second mass \( m_2 \) is labeled as 33.7 kg and is positioned on an incline. - The incline is at an angle \( \theta = 39.1^\circ \). **Part A) What is the tension in the cable?** There is an equation provided for the cable tension: \[ F_T = \text{unit in newtons (N)} \] A text box is also available for inputting the value of the tension \( F_T \). The task involves calculating the tension in the cable connecting the two masses, considering the forces acting on the system.
Transcribed Image Text:The two masses shown below are connected by a cable and are **not moving**. Answer the following questions about the system. The diagram shows two masses connected by a cable over a pulley. - The first mass \( m_1 \) is labeled as 25.1 kg and is hanging vertically. - The second mass \( m_2 \) is labeled as 33.7 kg and is positioned on an incline. - The incline is at an angle \( \theta = 39.1^\circ \). **Part A) What is the tension in the cable?** There is an equation provided for the cable tension: \[ F_T = \text{unit in newtons (N)} \] A text box is also available for inputting the value of the tension \( F_T \). The task involves calculating the tension in the cable connecting the two masses, considering the forces acting on the system.
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