A person drags a 10kg package along by applying a force of 40N at a 30 degree angle.
a. Calculate the net force that will accelerate the package.
b. Calculate the horizontal acceleration in m/s of the package.
c. Calculate the magnitude of the force applied by the floor.
d. Give a reason why the magnitude is slightly less than the weight of the package.
Answer (2)
The net force accelerating the package is approximately 34.64 N, leading to a horizontal acceleration of about 3.464 m/s². The normal force exerted by the floor is 78.1 N, which is less than the weight of the package due to the vertical component of the applied force counteracting some of the weight. This configuration highlights the balance of forces in dynamics and the role of components in motion.
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Let's solve the problem step-by-step.
a. Calculate the net force that will accelerate the package.
The force applied is at an angle, so we need to resolve it into horizontal and vertical components.
Resolve the horizontal component (F_horizontal):
F h or i zo n t a l = F × cos ( θ )
Where:
F = 40 N (the applied force)
θ = 3 0 ∘
F h or i zo n t a l = 40 × cos ( 3 0 ∘ )
F h or i zo n t a l ≈ 40 × 0.866 = 34.64 N
Resolve the vertical component (F_vertical):
F v er t i c a l = F × sin ( θ )
F v er t i c a l = 40 × sin ( 3 0 ∘ )
F v er t i c a l ≈ 40 × 0.5 = 20 N
The net force that accelerates the package is the horizontal component:
F n e t = F h or i zo n t a l = 34.64 N
b. Calculate the horizontal acceleration in m/s² of the package.
Using Newton's second law, F = ma , solve for acceleration a :
a = m F n e t
where m = 10 kg .
a = 10 34.64 = 3.464 m/s 2
c. Calculate the magnitude of the force applied by the floor.
The force applied by the floor is the normal force, which counteracts the weight of the package and the vertical component of the applied force.
Weight of the package (W):
W = m × g = 10 × 9.8 = 98 N
Net force applied vertically (F_vertical):
The total upward force is balanced by the normal force, N , minus the vertical component of the applied force:
N = W − F v er t i c a l
N = 98 − 20 = 78 N
d. Reason why the magnitude of the force applied by the floor is slightly less than the weight of the package.
The floor applies a normal force which acts to balance the weight of the package and the vertical component of any other forces acting upon it. Here, the vertical component of the applied force (20 N upwards) reduces the normal force needed to support the package completely. Thus, the normal force is the weight of the package minus this vertical component.
