Is it easier to push or to pull

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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?

Free Body Diagram Forces to consider y-axis Handle Push/Pull x-axis x a a a

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

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

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

Push - FBD Handle Py x a a a Px Lawnmower mg Force of

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

Push – Newton’s nd 2 Law Sum of Forces in the x-direction = Px

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

Pull - FBD Handle Py x a a a Px Lawnmower mg Force of

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

Pull – Newton’s nd 2 Law Sum of Forces in the x-direction = Ff

Pull – Newton’s nd 2 Law Sum of Forces in the x-direction = Ff – Px = u. FN – 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 (5) Solving for FN: Substitute (2) into (1) FN = mg - P sin(a) x Sum x = -P cos(a) + u(mg - P sin(a)) Lawnmower Set equal to 0 0 = -P cos(a) + u(mg - P sin(a)) P(cos(a) + u sin(a)) = umg

Summary Calculations From our earlier calculations, we recall that at equilibrium (just before the

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.

Backup Slides •

Backup Slides •

Free Body Diagram Forces to consider y-axis Push/Pull Handle x-axis x a a a

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

Sum of Forces in the x-direction = Px – Ff = P cos(a) –

Sum of Forces in the x-direction = Px – Ff = P cos(a) – u. FN Push - FBD (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) Handle Py 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)) x a a a Px Lawnmower mg Force of Friction (Ff) Force Normal (FN)

Sum of Forces in the x-direction = Ff - Px = u. FN -

Sum of Forces in the x-direction = Ff - Px = u. FN - P cos(a) Pull - FBD (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) Handle Py 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)) x a a a Px Lawnmower mg Force of Friction (Ff) Force Normal (FN)