CHAPTER 4

FORCE AND MOTION

15. F₁ = (5.5 N)[(cos 30°) ^- (sin 30°) y] = (4.76 N) + (^-2.75 N) ,

F₂ = (3.5 N)[(cos 37°) + (sin 37°) y] = (2.80 N) + (2.11 N) .

SF = F₁ + F₂ + F₃ = 0, _F F₃ = ^-(F₁ + F₂) = (^-7.6 N) + (0.64 N) .

So the x-component of F₃ is .

35. The friction force is opposite the horizontal force. a = = = .

40. First find acceleration from kinematics. v_o = 0, v = 35 m/s, x = 0.50 m.

Taking x_o = 0.

v² = v2o + 2a(x - x_o), _F a = v EQ \o(\s\up4(2o = = 1.23 ´ 10³ m/s².

SF = ma = (0.056 kg)(1.23 ´ 10³ m/s²) = .

50. (a) For Jane: T = ma = (50 kg)(0.92 m/s²) = 46 N.

The force on John is also 46 N by Newton’s third law.

a_John = = toward Jane.

(b) x_John = a_John t² , x_Jane = a_Jane t².

So = = = 1.2.

Also x_John + x_Jane = 10 m, or x_John + 1.2 x_John = 10 m.

Therefore x_John = .

59. (a) SF_x = F cosq = (30 N) cos 37^° = 23.96 N = ma_x,

so a_x = _{= .}

(b) SF_y = N ^- F sinq ^- w = ma_y = 0,

so N = w + F sin 37^° = (25 kg)(9.80 m/s²) + (30 N) sin 37^° = .

64. (a) There are two rings. SF_y = T + T ^- w = 0.

So T = = = = .

(b) In the vertical direction: SF_y = T sinq + T sinq ^- mg = 0.

So T = = = .

67. The free-body diagrams of the three objects.

For m₁: T₁ ^- m₁ g = m₁ a, (1)

For m₃: T₂ ^- T₁ = m₃ a, (2)

For m₂: m₂ g ^- T₂ = m₂ a, (3)

(1) + (2) + (3) gives (m₂ ^- m₁)g = (m₁ + m₂ + m₃)a,

so a = .

(a) a = = .

(b) a = = ^-2.0 m/s². So it is .

73. (a) For m₁: SF_x = T ^- m₁ g sinq = m₁a_x. (1)

For m₂: SF_y = m₂ g ^- T = m₂ a_y. (2)

a_x = a_y = a in magnitude. (1) + (2) results (m₂ ^- m₁ sinq)g = (m₁ + m₂)a.

a = _{g =} ^´ (9.80 m/s²) _{= (}m₁ up and m₂ down)_.

_{(b) From (2) in (a), T =} m₂ (g - a) = (2.5 kg)(9.80 m/s^{2 -} 1.24 m/s²) = .

82. (a) SF_y = N ^- mg = ma_y = 0, _F N = mg.

SF_x = F ^- f_s = ma_x = 0 (on the verge of moving), so f_smax = m_s N = F,

or m_s = = = .

(b) Similarly, m_k = = .

83. Refer to the diagram in Exercise 4.82.

F = f_smax = m_s N = m_s mg = 0.69(40 kg)(9.80 m/s²) = .

93. SF_y = N ^- mg cosq = ma_y = 0, _F N = mg cosq.

SF_x = mg sinq ^- f_k = ma_x.

Or mg sinq ^- m_k N = mg sinq ^- m_k (mg cosq) = ma_x.

So m_k = = = .

97. For m₁: SF = T₁ ^- m₁ g = 0, (1)

For m₃: SF = T₂ ^- T₁ ^- f_smax = 0, (2)

For m₂: SF = m₂ g ^- T₂ = 0, (3)

(1) + (2) + (3) gives (m₂ ^- m₁)g ^- f_s = 0,

or (m₂ ^- m₁)g ^- m_s N₃ = (m₂ ^- m₁)g ^- m_s m₃ g = 0,

so m_s = = = .

Alternate method (more conceptual).

Since a = 0, T₁ = m₁ g and T₂ = m₂ g.

So for m₃, m₂ g ^- m₁ g ^- f_s = 0, or m₂ g ^- m₁ g ^- m_s m₃ g = 0.

Therefore m_s = .

99. (a) First assume that m₁ has the tendency to move up the incline.

For m₁:

SF_y = N₁ ^- m₁ g cosq = 0, _F N₁ = m₁ g cosq.

SF_x = T ^- m₁ g sinq ^- f_smax = T ^- m₁ g sinq ^- m_s N₁

= T ^- m₁ g sinq ^- m_s m₁ g cosq = 0.

So T = m₁ g sinq + m_s m₁ g cosq = m₁ g(sinq + m_s cosq).

For m₂:

SF_x = m₂ g ^- T = 0, _F m₂ g = T = m₁ g(sinq + m_s cosq).

Therefore m₂ = m₁ (sinq + m_s cosq) = (2.0 kg)(sin 37^° + 0.30 cos 37^°) = 1.7 kg.

Next assume the other extreme, that m₁ has the tendency to move down the incline.

The static friction force will now be pointing up the incline, so the term m_s m₁ g cosq becomes negative. Repeating the calculation: m₂ = m₁ (sinq ^- m_s cosq) = (2.0 kg)(sin 37^{° -} 0.30 cos 37^°) = 0.72 kg.

Therefore m₂ can be anywhere .

(b) When both are moving at constant velocity, the acceleration is still zero. However, m_s is replace by m_k.

Assume that m₁ has the tendency to move up the incline.

m₂ = m₁ (sinq + m_k cosq) = (2.0 kg)(sin 37^° + 0.20 cos 37^°) = 1.5 kg.

Assume that m₁ has the tendency to move down the incline.

m₂ = m₁ (sinq ^- m_k cosq) = (2.0 kg)(sin 37^{° -} 0.20 cos 37^°) = 0.88 kg.

Therefore m₂ can be anywhere between .