HW3: Forces

3

when the elevator is accelerating upward (+)

F = ma

N - mg = m * a

a = 1/3 g

Sup = N = 1/3mg + mg = 4/3 mg

4

T₂ = 8/9m₂g

What is T₂?
a = g/9 (descending)

1. Box 1 F = ma
   T₁ - m₁g - T₂ = -m₁a

2. Box 2 F = ma
   T₂ - m₂g = -m₂a

Use the F=ma for box 2 because no answer choices have T₁ and T₁ is unknown.

T₂ - m₂g = -m₂a
T₂ - m₂g = -m₂(1/9)g
T₂ = m₂g -m₂(1/9)g= 9/9 m₂g- 1/9 m₂g
T₂ = 8/9m₂g

5. 

9. m₂F_a/ (m₁ + m₂)/k

Explanation:

1. Draw FBD for 1 m₁ and m₂

2. Write F = ma equations
   F₁ = m₁a
   F_a- F_sp-m₁g = +m₁a
   F₂ = m₂a
   F_sp-m₂g = +m₂a

3. Plug in Hooke’s law F_sp = kx
   F_a- kx-m₁g = +m₁a
   kx - m₂g = +m₂a

4. Add 2 equations together to solve for a (they are the same)
   F_a- m₁g- m₂g = m₁a + m₂a

5. Simplify
   F_a- g(m₁ + m₂) = a(m₁ + m₂)
   a = F_a- g(m₁ + m₂) / (m₁ + m₂)
   a = (F_a/ (m₁ + m₂) - g

6. Plug a in step 5 into Force equation 2 to solve for x
   F equation 2: kx - m₂g = +m₂a
   kx = m₂a + m₂g → m₂(a+g)
   x = m₂(a+g)/k
   x = m₂(((F_a/ (m₁ + m₂) - g)+g)/k
   x = m₂F_a/ (m₁ + m₂)/k

6

Explanation:

1. Set up F_x and F_y equations
   F_x = ma_x
   F_ax = ma_x
   F_y = ma_y
   N-mg- F_ay = 0 not useful

2. Find F_ax & F_ay
   F_ax = F_acosθ
   F_ay = F_asinθ not useful

3. Solve for F_a with F_x
   F_ax = F_acosθ → ma_x =F_acosθ
   F_a = ma_x / cosθ

7

51.94 N

Explanation

1. Draw FBD and write F =ma equation for the bundle (only in y direction)
   T-mg = ma

• arrange to solve for T
  T = ma + ma = m(a+g)

2. Underline unknowns
   T = m(a+g)

3. Solve for a using kinematics
   givens: v_o = 0, Δy = 1 @ t = 1.4 s, constant a
   v_f = v_o + at
   Δy = v_ot + ½at²
   1 = 0 + ½a(1.4)²
   a =2/(1.4)²

4. Plug a into the T equation
   T = m((2/(a.4)²)+g)

8-9

Explanations:

8) Find magnitude of resultant F

1. Find net F_x and F_y
   F_x = 832 - 371
   F_y = 644 - 251

2. use x and y components to find F
   F = ((F_y)²+ (F_x)²)^1/2

9) Find direction of the force with respect to the 832 N force

1. tan^-1(F_y/F_x)

10-11

10) 2

11) 6

Newton’s 3rd Law:

• every action in nature, there is an equal and opposite reaction

• Newton's third law states that for every action in nature, there is an equal and opposite reaction. This means that when one object exerts a force on a second object, the second object simultaneously exerts a force of equal magnitude and opposite direction on the first

12

13

Explanation:

1. Draw FBD & F=ma for both
   23 kg: m₁g - T₁ = m₁a
   8.7 kg: T₁ - m₂g = m₂a

2. Find rate of acceleration of the 2 when they pass eachother
   (a is equal)

3. Add the 2 equations to solve for a
   m₁g -m₂g = m₁a + m₂a
   g(m₁ -m₂) = a(m₁ + m₂)
   a = g(m₁ -m₂) / (m₁ + m₂)
   

14-15

14) 7

15) 2

Explanation

15) Read correctly Newton’s 3rd Law

16

1.40 m/s²

Explanation:

1. draw FBD and F equations
   no friction
   x: F_x-Fg_x = ma
   y: N-F_y-Fg_y = 0 not useful

2. Solve for a with x F=ma
   a = F_x-Fg_x / m
   a = 19cosθ-mgsinθ / 3

17

Explanation:

1. set up F = ma
   x: Fg_x - F_fk = ma
   y: N- Fg_y = 0

2. expand equations/simplify, underline unknowns
   x: mgsinθ - μ_kN = ma
   y: N- Fg_y = 0
   N = mgcosθ

3. plugin N and solve for μ_k
   μ_kN = mgsinθ - ma
   μ_k= mgsinθ - ma/(mgcosθ)

18-20

18)

19)

20)

Explanations:

18) What is the frictional force acting and on the mass?
frictional force is static bc the block is @ rest

1. Set up F = ma equations
   x: F_fs - Fg_x = 0
   y: N- Fg_y = 0

2. Solve for Ffs
   x: F_fs = Fg_x = mgsinθ
   y: N- Fg_y = 0 not useful

19) What is the largest angle which the incline can have so the mass doesn’t slide

1. Set up F = ma equations for if the block doesnt slide (a = 0)
   x: F_fs - Fg_x = 0
   y: N- Fg_y = 0

2. Rearrange & plugin
   x: F_fs = mgsinθ = Nμ_s
   y: N = mgcosθ

3. Plug in N into the F_x equation
   mgsinθ = (mgcosθ) μ_s

4. Simplify to get rid of the 2 θ
   (mgsinθ)/(mgcosθ) = μ_s
   tanθ = μ_s = 0.73
   tan^-1(0.73)

20) What is the acceleration of the block down the incline if the angle of incline is 41º?

1. Set up F = ma equation
   x: Fg_x - F_fs = ma
   y: N- Fg_y = 0

2. Expand, plugin, rearrange
   F_fk because it is moving
   Fg_x - F_f bc block is acclerating to the left direction (+)
   x: mgsinθ- Nμ_k = ma
   y: N = mgcosθ

3. Rearrange to solve for a
   a = (mgsinθ- (mgcosθ)μ_k )/m

Swing and Child Centripetal Force

Explanation:

1. Set up FBD and F_cy=ma_cy
   a_c = v_o² / r
   T₁ + T₁ - mg = ma_c
   2T₁ - mg = mv_o² / r

solve for v_o

_{no normal force acting on the girl}

Tension is first in the equation because centripetal acceleration is inwards

22

Centripetal

Explanation:

1. draw FBD & F = ma
   mg - N = ma_c
   mg - N = mv_o² / r

2. solve for N

23

change of x y coordinate system.

find x and y components of N instead of mg
no friction

1. Draw FBD and F=ma
   x: N_x - 0 = ma_c
   Nsinθ - 0 = ma_c
   Nsinθ - 0 = mv_o² / r
   y: N_y-mg = 0
   Ncosθ = mg

2. divide the two equations to find tanθ angle
   Nsinθ - 0 = mv_o² / r divide by
   Ncosθ = mg
   tanθ = v_o² /g
   tan^-1(v_o² /g)

24

centripetal force

converto mass g to kg
centripetal force = T_x
1. Draw FBD and F = ma
x: T_x = 0 = F_c
Tsinθ = 0 = F_c
y: Tcosθ - mg = 0
2. Solve for T
Tcosθ - mg = 0
3. find x component of T