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

Free Body Diagram

Lawnmower

Push/Pull

aa

mg Force Normal (FN)

a

Handlexx-axis

Force of Friction (Ff)

y-axis

Forces to consider

Resolve the “Push/Pull” Vector

Push/Pull (P)

a

Py

Px

P

Push Px = +P cos(a) Py = -P sin(a)

Pull Px = -P cos(a) Py = +P sin(a)

Push - FBD

Lawnmower

aa

mg Force Normal (FN)

a

Handlex

Force of Friction (Ff)

Py

Px

Push – Newton’s 2nd Law

x

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)

Pull - FBD

Lawnmower

aa

mg Force Normal (FN)

a

Handlex

Force of Friction (Ff)

Py

Px

Pull – Newton’s 2nd Law

Lawnmower

x

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)

Summary CalculationsFrom 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

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Free Body Diagram

Lawnmower

Push/Pull

aa

mg Force Normal (FN)

a

Handlex

x-axis

Force of Friction (Ff)

y-axis

Forces to consider

Push - FBD

Lawnmower

aa

mg Force Normal (FN)

a

Handlex

Force of Friction (Ff)

Py

Px

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))

Pull - FBD

Lawnmower

aa

mg Force Normal (FN)

a

Handlex

Force of Friction (Ff)

Py

Px

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))