A dock worker drags a crate with a rope. The rope's tension on the crate is 90 N. Friction from the floor acts backward on it at 30 N. The crate's mass is 12 kg.
How large is the crate's acceleration?
You have seen that increasing the net force increases the acceleration in proportion, and that increasing the mass decreases it. One equation ties the two together and answers the crate question exactly.
Newton's second law as an equation:
a = F_net/m
The symbol F_net is the net external force on the object: every force acting on the object, combined into one. The whole compound F_net is the symbol.
F_net is never a single force by itself while other forces act.
For the crate, the tension and the friction act in opposite directions, so they combine by subtraction: F_net = 90 N − 30 N = 60 N, pointing the tension's way.
The 90 N tension alone is not F_net. Feeding a single force into the equation while others act is the classic slip.
In the equation a = F_net/m, what is F_net?
With F_net in newtons and m in kilograms, the acceleration comes out in meters per second squared.
A rope tows a sled with 50 N of tension while friction acts backward on the sled at 18 N. Enter F_net on the sled, in N.
The equation says in symbols what the last two lessons showed in tables: a grows in proportion to F_net, and shrinks as m grows.
Nils says: in a = F_net/m, when two forces act on an object, F_net is their combination, not either force alone. Is he correct?
Find the crate's acceleration: the rope's tension on it is 90 N, friction acts backward at 30 N, and its mass is 12 kg.
Write down the values:
tension = 90 N forward, friction = 30 N backward, m = 12 kg, a = ?
Find the net force first. Opposite directions combine by subtraction:
F_net = 90 − 30 = 60 N, forward
Write down the equation:
a = F_net/m
Substitute in the values, and calculate:
a = 60 ÷ 12 = 5.0
The crate's acceleration is 5.0 m/s², in the direction of the tension.
Two ropes tow a sled, with tensions of 40 N and 25 N in the same direction, while friction acts backward at 15 N. What is F_net on the sled?
Forces in the same direction add; a force in the opposite direction subtracts.
F_net = 40 + 25 − 15 = 50
F_net = 50 N, in the ropes' direction.
Which statement is Newton's second law?
A mover drags a box: the rope's tension on it is 75 N, and friction acts backward at 27 N. Enter F_net on the box, in N.
Which set pairs each symbol in a = F_net/m with its unit?