PHYSICS 1: Forces
Kalamazoo Valley Community College
A box is pulled to the right by two forces F1 and F2 and to the left by F3.
What is the magnitude of F2 if the pulls and pushes need to balance?
F3 = 15N
F1 = 10N
m
F2= ???
This should be an intuitive question. F2 must be 5N. Take a moment and
think how you came to 5N. You took the difference between F3 and F1
and knew the remainder had to be equal to F2.
Now try to place this into a mathematical statement.
𝐹3 = 𝐹1 + 𝐹2
15𝑁 = 10𝑁 + 𝐹2
𝐹2 = 5𝑁
Now we shall make a formal mathematical statement out of the intuitive
math from the page before.
𝐹𝑥 = 0
Add up all
Forces
In the x-direction
Try the example again using the above statement.
𝐹𝑥 = 0
Points left (-)
𝐹1 + 𝐹2 − 𝐹3 = 0
10𝑁 + 𝐹2 − 15𝑁 = 0
𝐹2 = 5𝑁
they equal zero
Example
A traffic light is suspended by two cables. One of
the wires is at 35.0° to the horizontal and the
other is at 50.0°. If the traffic light has a mass of
40.0kg solve for the tension in the cables.
55°
35°
Draw the traffic light with only the forces acting on it.
Each cable has a tension and we have to name the
forces. T1 and T2 will do.
T1
T2
55°
Next we will make a table to organize the
forces before we add them up. List all the
forces on the object.
35°
X-direction
W = mg
Y-direction
T1x
T1y
T2x
T2y
Wx
Wy
We need to separate the forces into
their x and y components. Then fill in
each blank in the table. We have to fill
in every spot.
𝑇1𝑥 = 𝑇1 ∙ 𝑐𝑜𝑠 35°
T1
T1y
35°
𝑇1𝑥 = 0.819 ∙ 𝑇1
𝑇1𝑦 = 𝑇1 ∙ 𝑠𝑖𝑛 35°
T1x
𝑇1𝑦 = 0.574 ∙ 𝑇1
𝑇2𝑥 = 𝑇2 ∙ 𝑐𝑜𝑠 125°
T2
T2y
𝑇2𝑥 = −0.574 ∙ 𝑇2
55°
125°
𝑇2𝑦 = 𝑇2 ∙ 𝑠𝑖𝑛 125°
T2x
𝑇2𝑦 = 0.819 ∙ 𝑇2
-90°
W = mg
𝑊𝑥 = 𝑊 ∙ 𝑐𝑜𝑠 −90°
𝑊𝑥 = 0 ∙ 𝑊
𝑊𝑦 = 𝑊 ∙ 𝑠𝑖𝑛 −90°
𝑊𝑦 = −1 ∙ 40𝑘𝑔 ∙ 9.80𝑠𝑚2
X-direction
Y-direction
T1x
0.819 ∙ 𝑇1
T1y
0.574 ∙ 𝑇1
T2x
−0.574 ∙ 𝑇2
T2y
0.819 ∙ 𝑇2
Wx
0
Wy
-392N
Now we will add up all the forces in the x-direction.
𝐹𝑥 = 0
0.819 ∙ 𝑇1 + −0.574 ∙ 𝑇2 +0 = 0
0.819 ∙ 𝑇1 = 0.574 ∙ 𝑇2
0.819 ∙ 𝑇1 = 0.574 ∙ 𝑇2
𝑇1 = 0.700 ∙ 𝑇2
While we did not get a number for T1 or T2 we have a relationship we can use in the ydirection.
Next we will add up all the Forces in the y-direction and use the relationship from
the x-direction.
𝐹𝑦 = 0
0.574 ∙ 𝑇1 +0.819 ∙ 𝑇2 −392N = 0
𝑇1 = 0.700 ∙ 𝑇2
0.574 ∙ 0.700 ∙ 𝑇2 + 0.819 ∙ 𝑇2 −392N = 0
0.402 ∙ 𝑇2 +0.819 ∙ 𝑇2 −392N = 0
1.22 ∙ 𝑇2 = 392N
𝑇2 = 321𝑁
𝑇1 = 0.700 ∙ 321𝑁
𝑇1 = 225N