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Is it easier to push or to pull a lawnmower or vacuum cleaner?

Is it easier to push or to pull a lawnmower or vacuum cleaner?. Forces to consider. Free Body Diagram. y -axis. Push/Pull. Handle. x-axis. x. a. aa. Force of Friction ( Ff ). Lawnmower. m g. Force Normal (FN). Resolve the “Push/Pull” Vector. P. Push Px = +P cos(a)

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Is it easier to push or to pull a lawnmower or vacuum cleaner?

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  1. Is it easier to push or to pull a lawnmower or vacuum cleaner?

  2. Forces to consider Free Body Diagram y-axis Push/Pull Handle x-axis x a aa Force of Friction (Ff) Lawnmower mg Force Normal (FN)

  3. Resolve the “Push/Pull” Vector P Push Px = +P cos(a) Py = -P sin(a) Pull Px = -P cos(a) Py = +P sin(a) Push/Pull (P) Py a Px

  4. Push - FBD Handle Py x a aa Force of Friction (Ff) Px Lawnmower mg Force Normal (FN)

  5. Push – Newton’s 2nd Law Sum of Forces in the x-direction = Px – Ff = P cos(a) – uFN (1) Sum of Forces in the y-direction = FN – Py – mg = FN – P sin(a) - mg = 0 for a body at rest Solving for FN: FN = mg +P sin(a) (2) Substitute (2) into (1) Sum x = P cos(a) – u(mg + P sin(a)) Set equal to 0 0 = P cos(a) – u(mg + P sin(a)) P(cos(a) – u sin(a)) = umg Or Push = umg/(cos(a)-usin(a)) (3) x

  6. Pull - FBD Handle Py x a aa Force of Friction (Ff) Px Lawnmower mg Force Normal (FN)

  7. Pull – Newton’s 2nd Law Sum of Forces in the x-direction = Ff– Px = uFN – P cos(a) (4) Sum of Forces in the y-direction = FN + Py – mg = FN + P sin(a) - mg = 0 for a body at rest Solving for FN: FN = mg - P sin(a) (5) Substitute (2) into (1) Sum x = -P cos(a) + u(mg - P sin(a)) Set equal to 0 0 = -P cos(a) + u(mg - P sin(a)) P(cos(a) + u sin(a)) = umg Or Pull = umg/(cos(a)+usin(a)) (6) x Lawnmower

  8. Summary Calculations From our earlier calculations, we recall that at equilibrium (just before the lawnmower moves) “Push” and “Pull” may be represented as shown below: Push = umg/(cos(a)-usin(a)) (3) Pull = umg/(cos(a)+usin(a)) (6) Now, we know that cos(a) and sin (a) are less than or equal to 1.00. Also, u is less than 1.00. As a result, the denominator of (3) will always be less than the denominator of (6). Therefore, the “Push” will always exceed “Pull”. The force required to push a lawnmower is great than the force required to pull it.

  9. Backup Slides

  10. Forces to consider Free Body Diagram y-axis Push/Pull Handle x-axis x a aa Force of Friction (Ff) Lawnmower mg Force Normal (FN)

  11. Sum of Forces in the x-direction = Px – Ff = P cos(a) – uFN (1) Sum of Forces in the y-direction = FN – Py – mg = FN – P sin(a) – mg = 0 for a body at rest Solving for FN: FN = mg +P sin(a) (2) Substitute (2) into (1) Sum x = P cos(a) – u(mg + P sin(a)) Set equal to 0 0 = P cos(a) – u(mg + P sin(a)) P(cos(a) – u sin(a)) = umg Or Push = umg/(cos(a)-usin(a)) Push - FBD Handle Py x a aa Force of Friction (Ff) Px Lawnmower mg Force Normal (FN)

  12. Sum of Forces in the x-direction = Ff - Px = uFN - P cos(a) (1) Sum of Forces in the y-direction = FN + Py – mg = FN + P sin(a) – mg = 0 for a body at rest Solving for FN: FN = mg - P sin(a) (2) Substitute (2) into (1) Sum x = -P cos(a) + u(mg - P sin(a)) Set equal to 0 0 = -P cos(a) + u(mg - P sin(a)) P(cos(a) + u sin(a)) = umg Or Pull = umg/(cos(a)+usin(a)) Pull - FBD Handle Py x a aa Force of Friction (Ff) Px Lawnmower mg Force Normal (FN)

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