Engineering Mechanics - Force Vectors - Discussion
Discussion Forum : Force Vectors - General Questions (Q.No. 5)
5.
Force F acts on peg A such that one of its components, lying in the x-y plane, has a magnitude of 50 lb. Express F as a Cartesian vector.
Discussion:
7 comments Page 1 of 1.
Anonymous said:
6 years ago
See, F is making 60 degrees with the z-axis, but the question says "one of its components lying in the XY axis has a magnitude of 50". Therefore, we consider F making 30 degrees with X-axis (90-60 = 30) which gives => 50 = FCos30.
Anonymous said:
6 years ago
How come everyone see 30 degree? I only see 60 degree only.
Manoj said:
9 years ago
We know that x component of the force F is F(x) = Fcos30.
From that F = F(x)/cos30= 57.77
From the figure, we come to know that the direction of the force is in +ve x direction -ve y direction and +ve z direction.
Then take the two options which have negative j.
Ffind z component F(y)k= Fcos60 z = 57.7*cos60=28.86k.
Hence Option D is correct.
From that F = F(x)/cos30= 57.77
From the figure, we come to know that the direction of the force is in +ve x direction -ve y direction and +ve z direction.
Then take the two options which have negative j.
Ffind z component F(y)k= Fcos60 z = 57.7*cos60=28.86k.
Hence Option D is correct.
GERSHOM NTSAKO said:
9 years ago
50 = Fcos30.
F = 57.74.
57.74sin(30) = Z-COMPONENT.
= 28.867.
THEREFORE x-component should be +ve and y-component will be -ve.
F = 57.74.
57.74sin(30) = Z-COMPONENT.
= 28.867.
THEREFORE x-component should be +ve and y-component will be -ve.
Ameet Kumar said:
10 years ago
Kindly sir please give me the specific solution?
Riyaz Ibrahim said:
10 years ago
Go through F = 50sin60-50cos60+50tan60.
Aimen said:
1 decade ago
Please give me the solution.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers