Tension word problems
To solve these tension word problems, you will mostly need Newton's second law of motion.
Problem # 1
a 100 kg bucket is being lifted by a rope.
The bucket started at rest and after being lifted 4 m, it is moving at 4
m/s. If the acceleration is constant, what is the tension on the rope?
The diagram above describes the problem and it is useful in order to visualize the problem.
is the force of tension and it is represented with a bigger vector. This force has to be bigger since it is making the bucket go up.
The bucket started from rest, so V0
Then the bucket moved with a speed of 4 m/s, so V = 4 m/s
is the force of gravity
The positive direction is chosen to be up
= m × a + m × g
= m × (a + g)
The only missing quantity is a. To get a, we can use one of the constant acceleration equations.
Since we know V0
, V, and d, we can use V2
+ 2 × a × 4
16 = 0 + 8a
16 = 8a
a = 2 m/s2
= 100 × (2 + 9.8)
= 100 × 11.8 = 1180 Newtons
Problem # 2:
is attached to a vehicle weighing 500 kN. A very strong man puts the
other end of the rope inside his mouth at an angle of 30 degrees from
the horizontal. If the strong man is able to move the vehicle with a
constant force 1 m away with a speed of 0.25 m/s, what is the tension on
the rope. (Use g = 10 m/s2)
This situation is shown below with graph.
Newton's second law gives F = ma
Since the vehicle is only moving along the x axis, the only force on the vehicle is along the x axis.
We get Fx
The force Fx
that we need here is the horizontal component of the tension force.
= T cos(30°)
T cos(30°) = max
We need to find the acceleration and the mass of the vehicle.
= 50000 kg
Since the man is moving the vehicle with a constant force, the
acceleration is constant. Therefore, we can use the following equation
again to get the acceleration.
+ 2 × a × 1
0.0625 = 0 + 2a
0.0625 = 2a
a = 0.03125 m/s2
We can now find the tension on the rope
T cos(30°) = max
T cos(30°) = 50000 × 0.03125
T cos(30°) = 1562.5
T × 0.866 = 1562.5
T = 1562.5 / 0.866
T = 1804.27 Newtons
Tension word problems and constant acceleration equations
Notice how important it was to use one of the constant acceleration equations to solve the two tension word problems above. Many problems in physics have a constant acceleration. Therefore, try to become familiar to these equations since they will keep coming back when solving word problems.
Nov 18, 20 01:20 PM
Top-notch introduction to physics. One stop resource to a deep understanding of important concepts in physics
New math lessons
Your email is safe with us. We will only use it to inform you about new math lessons.