Tony Pulls Wagon: Work Problem Solutions

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Homework Help Overview

The problem involves calculating the work done by Tony as he pulls a wagon at an angle. The context is centered around the concepts of work, force, and the impact of angles in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relevance of the angle of the force applied to the work done, questioning whether the angle affects the amount of work. Some suggest resolving the force into its components to understand its impact on work calculation.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of how the angle influences work. Some guidance has been offered regarding the resolution of forces into components and the use of the dot product in calculating work.

Contextual Notes

There is a focus on the horizontal component of the force and its relationship to the displacement, with some participants noting that the wagon does not gain elevation, which may affect the normal force consideration.

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Tony pulls his wagon a distance of 15 m across the garden while applying a force of 160 N on the wagon’s handle. If the handle makes an angle of 40degree with the horizontal, how much work did Tony do on the wagon?

I have an idea that W= F*D but why does question have the fact that handle makes an angle horizantal 40 degree angle?...
 
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Because one is looking for the work applied in the direction of travel (and parallel with the horizontal). The force at an angle has horizontal and vertical components. Only the force parallel to travel is doing the work. The normal force is static since the wagon is not gaining elevation in the gravity field.
 
so degrees doesn't affect the amount of work?...
 
does anyone know this problem??
 
You have to resolve the force into vertical and horizontal components. the component that is in the same direction as the force you call that [itex]F_1[/itex] and then you use the formula [itex]W=F_1s[/itex] where s is the displacement IN THE DIRECTION OF THE FORCE.

yes the angle at which the force acts varies that amount of work that it does
 
rhxoehwhfh said:
so degrees doesn't affect the amount of work?...
Resolve the force into vertical and horizontal components.

Using the angle [itex]\theta[/itex] with respect to horizontal, FH = F cos[itex]\theta[/itex], which is the force doing the work,

and vertical, FV = F sin[itex]\theta[/itex]

or x (horizontal) and y (vertical).

If one does F*d, where * implies the dot product of the force vector and the displacement vector, then the cos of the angle is part of the solution. The dot product means one is applying the component of force parallel with the direction of displacement.
 
Last edited:
thank you
 

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