Answer :
A. The acceleration of the golf bag is approximately 4.56 m/s².
B. The push force on the golf bag is approximately 13.88 lb.
Given:
Mass of golfer (m_g) = 199 lb ≈ 90.34 kg (converted to kg for calculations)
Mass of golf bag (m_b) = 20 lb ≈ 9.07 kg
Angle of incline (θ) = 13°
Tractive force (F_t) = 93 lb ≈ 413.69 N
Acceleration due to gravity (g) = 9.81 m/s² (approximated)
Assumptions:
The golfer pushes the bag parallel to the ground.
The wheels of the golf bag roll freely (no friction).
We neglect air resistance.
A. Determine the acceleration of the golf bag:
To find the acceleration of the golf bag, we need to find the net force acting on it. We'll resolve the forces along the x-axis (parallel to the ground) and y-axis (perpendicular to the ground).
Forces on the golf bag:
Push force (F_p) applied by the golfer (unknown, to be found in part B)
Force due to gravity (F_g) = m_b * g * sin(θ) = 9.07 kg * 9.81 m/s² * sin(13°) ≈ 20.45 N (opposing motion)
Normal force (N) = m_b * g * cos(θ) ≈ 87.31 N (perpendicular to motion, not affecting acceleration)
Net force along x-axis:
F_net = F_p - F_g
We know the tractive force (F_t) is equal to the force required to move the golfer and the golf bag up the hill. Since the wheels roll freely, the tractive force is equal to the sum of the forces opposing motion:
F_t = F_g * (m_g + m_b) / m_g
Now, we'll use the tractive force to find the acceleration:
F_net = F_t * (m_b / (m_g + m_b)) = 413.69 N * (9.07 kg / (90.34 kg + 9.07 kg)) ≈ 41.38 N
Acceleration (a) = F_net / m_b = 41.38 N / 9.07 kg ≈ 4.56 m/s²
B. Determine the push force on the golf bag:
Now that we have the acceleration, we can find the push force:
F_p = F_net + F_g = 41.38 N + 20.45 N ≈ 61.83 N
To convert this to pounds:
F_p ≈ 61.83 N * (1 lb / 4.45 N) ≈ 13.88 lb
