Engineering Mechanics - Force Vectors - Discussion

Discussion Forum : Force Vectors - General Questions (Q.No. 11)
11.

Express the force F1 in Cartesian vector form.

F1 = (200 i - 200 j + 283 k) lb
F1 = (200 i + 200 j + 283 k) lb
F1 = (-200 i + 200 j + 565 k) lb
F1 = (-200 i + 200 j + 283 k) lb
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
7 comments Page 1 of 1.

Afifah said:   5 years ago
Can someone show me how to find α and β for F2?

I got the same problem whenever the only angle on the x-y plane is given, not directly from x and y-axis. Thank you in advance!

Laxman said:   6 years ago
Thanks @Manoj.

Manoj said:   8 years ago
f = xi + yj + zk.

x is x component of force
y is y component of force
z is z component of force

By Considering the direction of the force
It is making 60 deg with positive y-axis and negative x-axis. and 45 deg with positive z-axis.
Cos theta(x) = x component of force/ magnitude of x.

Therefore x component = -(cos 60*400) = 200 since it is in negative direction
Similarly y component = cos 60 * 400 = 200.
z component = cos 45 * 400 = 283.
Therefore force in cartesian form is -200i+200j+283k.
(1)

Sourabh said:   9 years ago
Please give the complete solution of this question.

Vikash kumar said:   10 years ago
Please give me the complete solution of this question.

Sticki said:   1 decade ago
The answer is right. Take it as cos60 along the y-axis.

Preeti2009_delhi@yahoo.com said:   1 decade ago
Answer for this question should be A because cos120=-1/2 along with y-axis.

Please give me reply.

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