2D Forces: Pulling & Pushing

Review: Free Body Diagrams

• Free body diagrams are used to show the relative magnitude and direction of all forces acting on an object

But what if some of the forces are not in the x or y plane?

Forces at Angles

• We will break the forces with angles into their x and y components

Part 1: Pulling

F_N

F_g

F_f

F_px

F_p = pulling force

F_p

F_py

motion

y component

• F_g = mg
• sinθ = F_py(opp)/F_{p (hyp)}
• F_nety = 0
• F_nety = F_N + F_py + F_g

x component

• F_netx = ma
• cosθ = F_px(adj)/F_{p (hyp)}
• F_f = μF_N
• F_netx = 0

(if constant velocity)

• F_netx = F_px + F_f

Sample Problem 1

A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:

(a) F_N (b) F_f (c) F_netx

​

FIRST: Draw your FBD

SECOND: List your givens (go ahead and find F_px and F_py)

​

F_N

F_g

F_px

F_p = 70N

F_py

motion

x components y components both

μ = 0.4 F_nety = 0N θ = 35°

m = 15 kg

F_p = 70N

F_f

Sample Problem 1

A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:

(a) F_N

• F_nety = F_N + F_py + F_g

​

​

• F_nety = 0
• Need to find F_py and F_g

F_N

F_g

F_px

F_p = 70N

F_py

motion

F_f

In the previous unit, F_N = F_g, but because the applied force is in both the x and y direction, F_N ≠ F_g

F_g = mg

F_g = (15)(-9.8)

F_g = -147N

F_py = F_psinθ

F_pv = 70sin35

F_pv = 40N

0 = F_N + 40 + (-147)

F_N = 107N

Sample Problem 1

A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:

(a) F_N (b) F_f (c) F_netx

​

UPDATED GIVENS:

F_N

F_g

F_px

F_p = 70N

F_py

motion

x components y components both

μ = 0.4 F_nety = 0N θ = 35°

F_g = -147N m = 15 kg

F_N = 107N F_p = 70N

F_py = 40N

F_f

Sample Problem 1

A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:

(b) F_f

• F_f = μF_N
• F_f = (0.4)(107)
• F_f = -42.8N

F_N

F_g

F_px

F_p = 70N

F_py

motion

F_f

Sample Problem 1

A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:

(a) F_N (b) F_f (c) F_netx

​

UPDATED GIVENS:

F_N

F_g

F_px

F_p = 70N

F_py

motion

x components y components both

μ = 0.4 F_nety = 0N θ = 35°

F_f = 42.8N F_g = -147N m = 15 kg

F_N = 107N F_p = 70N

F_py = 40N

F_f

Sample Problem 1

A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:

(c) F_netx

• F_netx = F_px + F_f
• Need to find F_px

​

​

​

• F_netx = 57 + (-42.8)
• F_netx = 14.2N

F_N

F_g

F_px

F_p = 70N

F_py

motion

F_f

F_px = F_pcosθ

F_px = 70cos35

F_px = 57N

Sample Problem 2

• A 52-kg crate is pulled across the ice with a rope. A force of 125 N is applied to the rope at an angle of 37^o with the horizontal. The coefficient of friction between the crate and ice is 0.2. Calculate the following quantities:

(a) F_N (b) F_f (c) F_netx (d) a

Lesson Check 4.3

Part 2: Pushing

F_N

F_g

F_f

F_px

F_p = pushing force

F_p

F_py

y component

• F_g = mg
• sinθ = F_py(opp)/F_{p (hyp)}
• F_nety = 0
• F_nety = F_N + F_py + F_g

x component

• F_netx = ma
• cosθ = F_px(adj)/F_{p (hyp)}
• F_f = μF_N
• F_netx = 0

(if constant velocity)

• F_netx = F_px + F_f

F_py is negative due to the down direction

What’s the Difference?

• The difference between calculating for pushing instead of pulling is that the F_py force is negative

Sample Problem 1

A person pushes a 14 kg lawn mower at constant speed with a force of F = 88N directed along the handle, which is at an angle of 45° to the horizontal. Calculate

• the normal force exerted vertically upward on the mower by the ground
• the horizontal friction force on the mower

​

FIRST: Draw your FBD

SECOND: List your givens (go ahead and find F_px and F_py)

​

​

F_g

F_px

F_p = 88N

F_py

motion

F_f

F_N

x components y components both

F_netx = 0N F_nety = 0N θ = 45°

m = 14 kg

F_p = 88N

Sample Problem 1

A person pushes a 14 kg lawn mower at constant speed with a force of F = 88N directed along the handle, which is at an angle of 45° to the horizontal. Calculate

• the normal force exerted vertically upward on the mower by the ground

F_nety = F_N + F_py + F_g

F_nety = 0

​

Need to find F_py and F_g (both down, both negative)

​

​

​

​

0 = F_N + -62 + (-137.2)

F_N = 199.2N

F_g

F_px

F_p = 88N

F_py

motion

F_f

F_N

F_py = 88sin45

F_py = -62N

F_g = (14)(-9.8)

F_g = -137.2N

Sample Problem 1

A person pushes a 14 kg lawn mower at constant speed with a force of F = 88N directed along the handle, which is at an angle of 45° to the horizontal. Calculate

• the normal force exerted vertically upward on the mower by the ground
• the horizontal friction force on the mower

​

UPDATED GIVENS:

​

F_g

F_px

F_p = 88N

F_py

motion

F_f

F_N

x components y components both

F_netx = 0N F_nety = 0N θ = 45°

F_py = 62N m = 14 kg

F_g = -137.2N F_p = 88N

F_N = 199.2N

Sample Problem 1

A person pushes a 14 kg lawn mower at constant speed with a force of F = 88N directed along the handle, which is at an angle of 45° to the horizontal. Calculate

(b) the horizontal friction force on the mower

​

F_netx = F_px + F_f

F_netx = 0

​

Need to find F_px

​

​

​

​

​

0 = 62 + F_f

F_f = -62N

F_g

F_px

F_p = 88N

F_py

motion

F_f

F_N

F_px = 88cos45

F_px = 62N