How Do You Calculate the Forces Acting on Grandma and Her Chair?

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The discussion revolves around calculating the forces acting on Grandma and her chair, specifically addressing problems related to horizontal force and acceleration. Grandma, weighing 500N, applies a horizontal force of 100N to her 700N chair, resulting in a steady acceleration of 0.200 m/s². Key forces at play include the normal force and friction, which are essential for understanding the chair's movement. For calculating the resultant force of a traction device, the Pythagorean theorem is recommended to determine the magnitude, while the angle can be found using the inverse tangent of the components. Overall, the conversation emphasizes the importance of understanding vector components in solving these physics problems.
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Hello everyone, I am having some trouble with some problems. For the first picture I do not understand how to do #19. Please help me! Thank you!

Homework Statement



Grandma, of weight 500N, exerts a steady horizontal force of 100N on her 700N wing chair, thereby pushing chair across the parlor floor. While Grandma is pushing the chair, her feet do not slip on the floor, and both Grandma and the chair accelerate together at a steady 0.200 m/s^2.

Homework Equations


The Attempt at a Solution


IMAG0126.jpg


For the second picture, I do not know how to solve the last question on resultant force of the traction device that exerts on the foot. Please refer to picture:
IMAG0125.jpg


For my attempt, I came up with my x and y scalar components and tried adding them but nothing came out to be any of the values. I am just very confused with this problem.
 
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You're right for 19. There are two forces acting on the chair: the normal force and friction. We know there must be friction because the chair would have a higher acceleration otherwise.

For 25, I assume you've solved the other three parts (and so have your x- and y- components). Now, as with any vector whose components are known, use the Pythagorean theorem to get the magnitude of the resultant traction. To get its angle, take the inverse tangent of the y component over the x component.
 
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