Component Method to Add Vectors
In many situations, there are more than one external force acting on an object. Calculating the
net external force on the object can be obtained by using component method, law of sine and law
of cosine method, and graphical method.
Example 4
Consider two force vectors, 𝐴⃗ and 𝐵
⃗⃗
𝐴⃗ = 5.0𝑁, 340 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 + 𝑥 − 𝑎𝑥𝑖𝑠
⃗⃗ = 5.0𝑁, 1050 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 + 𝑥 − 𝑥𝑖𝑠.
𝐵
Using component method, determine the resultant vector 𝑅⃗⃗ = 𝐴⃗ + 𝐵
⃗⃗
a) Expressed in unit vector notation 𝑖⃗, 𝑗⃗, 𝑘⃗⃗ and
b) resultant’s magnitude and direction relative to the +x-axis.
Solution:
𝐴⃗ = 5.0𝑁, 340 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑡ℎ𝑒 + 𝑥 − 𝑎𝑥𝑖𝑠 ⃗⃗ = 5.0𝑁, 1050 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑡ℎ𝑒 + 𝑥 − 𝑥𝑖𝑠.
𝐵
y
⃗⃗
𝐵
𝐴⃗ ⃗⃗⃗⃗⃗
𝐵 𝑦
𝐴⃗𝑦
1050
340 x ⃗⃗⃗⃗⃗
𝐵𝑥
𝐴⃗𝑦
𝐴⃗ = 𝐴𝑥 𝑖⃗ + 𝐴𝑦 𝑗⃗ + 𝐴𝑧 𝑘⃗⃗ ⃗⃗ = 𝐵𝑥 𝑖⃗ + 𝐵𝑗⃗ + 𝐵𝑧 𝑘⃗⃗
𝐵
𝐴⃗ = (5.0𝑁𝑐𝑜𝑠34𝑜 )𝑖⃗ + (5.0𝑁𝑠𝑖𝑛34𝑜 )𝑗⃗ + 0𝑘⃗⃗ ⃗⃗ = (5.0𝑁𝑐𝑜𝑠105𝑜 )𝑖⃗ + (5.0𝑁𝑠𝑖𝑛105𝑜 )𝑗⃗ + 0𝑘⃗⃗
𝐵
𝐴⃗ = 4.1𝑁 𝑖⃗ + 2.8𝑁 𝑗⃗ ⃗⃗ = −1.3𝑁 𝑖⃗ + 4.8𝑁 𝑗⃗
𝐵
a) 𝑅⃗⃗ = 𝐴⃗ + 𝐵
⃗⃗ =𝑅𝑥 𝑖⃗ + 𝑅𝑦 𝑗⃗=(𝐴𝑥 + 𝐵𝑥 )𝑖⃗ + (𝐴𝑥 + 𝐵𝑥 )𝑗⃗ = (𝐴𝑥 + 𝐵𝑥 )𝑖⃗ + (𝐴𝑥 + 𝐵𝑥 )𝑗⃗
= (4.1𝑁 − 1.3𝑁)𝑖⃗ + (2.8𝑁 + 4.8𝑁)𝑗⃗
𝑅⃗⃗ = 2.8𝑁 𝑖⃗ + 7.6𝑁 𝑗⃗
𝑅⃗⃗ b) Magnitude of 𝑅⃗⃗
𝑅𝑦 𝑗⃗
𝑅 = √𝑅𝑥 2 + 𝑅𝑦 2
𝛼
𝑅 = √(2.8𝑁)2 + (7.6𝑁)2
𝑅𝑥 𝑖⃗
𝑅 = 8.1 𝑁
In many situations, there are more than one external force acting on an object. Calculating the
net external force on the object can be obtained by using component method, law of sine and law
of cosine method, and graphical method.
Example 4
Consider two force vectors, 𝐴⃗ and 𝐵
⃗⃗
𝐴⃗ = 5.0𝑁, 340 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 + 𝑥 − 𝑎𝑥𝑖𝑠
⃗⃗ = 5.0𝑁, 1050 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 + 𝑥 − 𝑥𝑖𝑠.
𝐵
Using component method, determine the resultant vector 𝑅⃗⃗ = 𝐴⃗ + 𝐵
⃗⃗
a) Expressed in unit vector notation 𝑖⃗, 𝑗⃗, 𝑘⃗⃗ and
b) resultant’s magnitude and direction relative to the +x-axis.
Solution:
𝐴⃗ = 5.0𝑁, 340 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑡ℎ𝑒 + 𝑥 − 𝑎𝑥𝑖𝑠 ⃗⃗ = 5.0𝑁, 1050 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑡ℎ𝑒 + 𝑥 − 𝑥𝑖𝑠.
𝐵
y
⃗⃗
𝐵
𝐴⃗ ⃗⃗⃗⃗⃗
𝐵 𝑦
𝐴⃗𝑦
1050
340 x ⃗⃗⃗⃗⃗
𝐵𝑥
𝐴⃗𝑦
𝐴⃗ = 𝐴𝑥 𝑖⃗ + 𝐴𝑦 𝑗⃗ + 𝐴𝑧 𝑘⃗⃗ ⃗⃗ = 𝐵𝑥 𝑖⃗ + 𝐵𝑗⃗ + 𝐵𝑧 𝑘⃗⃗
𝐵
𝐴⃗ = (5.0𝑁𝑐𝑜𝑠34𝑜 )𝑖⃗ + (5.0𝑁𝑠𝑖𝑛34𝑜 )𝑗⃗ + 0𝑘⃗⃗ ⃗⃗ = (5.0𝑁𝑐𝑜𝑠105𝑜 )𝑖⃗ + (5.0𝑁𝑠𝑖𝑛105𝑜 )𝑗⃗ + 0𝑘⃗⃗
𝐵
𝐴⃗ = 4.1𝑁 𝑖⃗ + 2.8𝑁 𝑗⃗ ⃗⃗ = −1.3𝑁 𝑖⃗ + 4.8𝑁 𝑗⃗
𝐵
a) 𝑅⃗⃗ = 𝐴⃗ + 𝐵
⃗⃗ =𝑅𝑥 𝑖⃗ + 𝑅𝑦 𝑗⃗=(𝐴𝑥 + 𝐵𝑥 )𝑖⃗ + (𝐴𝑥 + 𝐵𝑥 )𝑗⃗ = (𝐴𝑥 + 𝐵𝑥 )𝑖⃗ + (𝐴𝑥 + 𝐵𝑥 )𝑗⃗
= (4.1𝑁 − 1.3𝑁)𝑖⃗ + (2.8𝑁 + 4.8𝑁)𝑗⃗
𝑅⃗⃗ = 2.8𝑁 𝑖⃗ + 7.6𝑁 𝑗⃗
𝑅⃗⃗ b) Magnitude of 𝑅⃗⃗
𝑅𝑦 𝑗⃗
𝑅 = √𝑅𝑥 2 + 𝑅𝑦 2
𝛼
𝑅 = √(2.8𝑁)2 + (7.6𝑁)2
𝑅𝑥 𝑖⃗
𝑅 = 8.1 𝑁