Q: Resolve a weight of 10 N in two directions which are parallel and perpendicular to a slope inclined at 30∘ to the horizontal.

Component perpendicular to the planew⊥=wcos30∘=(10)23=53 Nand component parallel to the planew∥∣=wsin30∘=(10)(21)=5 N

Q: Resolve horizontally and vertically a force F=8 N which makes an angle of 45∘ with the horizontal.

Horizontal component of F isFH=Fcos45∘=(8)(21)=42 Nand vertical component of F isFV=Fsin45∘=(8)(21)=42 NNote Two vectors given in the form of i^,j^ and k^ can be added, subtracted or multiplied (by a scalar) .

Q: the y component of a shown in the figure is 103. Find the magnitude of the vector. Also find its x component.

The y component is: y=acos30∘ ⇒⇒103a=a23=20.

Q: A vector is given as a=i^+j^+2k^ Find its magnitude and the angle that it makes with the x-axis.

Solution a=x2+y2+z2=12+12+(2)2=2 cosα⇒α=x2+y2+z2x=21=30∘.

Q: In a Cartesian coordinate system S, unit vectors along x and y axes are i^ and j^. A new coordinate system S1 is generated by rotating system S by an angle of 37∘ in the anticlockwise sense as shown in the figure.(i) Unit vectors in system S1 are i^1 and j^1. Express i^1 and j^1 in terms of i^ and j^.(ii) A vector in system S is expressed as r=2i^+2j^. Express the vector in system S1 using i^1 and j^1.[ Take sin37∘≃0.6]

(i) i^1=(1⋅cos37∘)i^+(1⋅sin37∘)j^ ⇒i^1=0.8i^+0.6j^ (a) and j^1=−(1⋅sin37∘)i^+(1⋅cos37∘)j^ ⇒j^1=−0.6i^+0.8j^ (b)

Q: Write vectors a,b,c and d using unit vectors i^,j^. Magnitudes are as indicated. [tan37∘=43]

Applying component of the vector a=52i^+52j^ b=−63i^+6j^ c=−6i^−8j^ d=3i^−4j^

Q: Find the component of b in the direction of a a=i^+j^b=2i^+3j^−k^

Component of b in the direction of a = ∣a∣a⋅bHence, 25

Q: F1,F2 and F3 are three forces, represented as shown. Each of them is of equal magnitude. P,Q,R and S represent various directions

Break F1 along x−y direction then apply addition and subtraction of vector. (c,d)

Q: For the situation shown below, which of the following results are correct?

Let force be F so resultant is in direction4i^+3j^+(5cos37∘i^+5sin37∘j^)+F=ki^⇒4i^+3j^+4i^+3j^+F=ki^⇒8i^+6j^+F=ki^⇒F=(k−8)i^−6j^⇒F=(k−8)2+(6)2⇒Fmin=6N

Question 1

Resolve a weight of 10 N in two directions which are parallel and perpendicular to a slope inclined at 30∘ to the horizontal.

Accepted Answer

Component perpendicular to the planew⊥​=wcos30∘=(10)23​​=53​ Nand component parallel to the planew∥∣​=wsin30∘=(10)(21​)=5 N

Question 2

Resolve horizontally and vertically a force F=8 N which makes an angle of 45∘ with the horizontal.

Accepted Answer

Horizontal component of F isFH​=Fcos45∘=(8)(2​1​)=42​ Nand vertical component of F isFV​=Fsin45∘=(8)(2​1​)=42​ NNote Two vectors given in the form of i^,j^​ and k^ can be added, subtracted or multiplied (by a scalar) .

Question 3

the y component of a shown in the figure is 103​. Find the magnitude of the vector. Also find its x component.

Accepted Answer

The y component is: y=acos30∘
⇒⇒​​103​a​=a23​​=20.​

Question 4

A vector is given as a=i^+j^​+2​k^ Find its magnitude and the angle that it makes with the x-axis.

Accepted Answer

Solution a=x2+y2+z2​=12+12+(2​)2​=2
cosα⇒α​=x2+y2+z2​x​=21​=30∘.​

Question 5

Geographical Co-ordinates Consider the centre of the Earth as the origin of a coordinate system. Rotation axis of the Earth is z-axis. The point (P) of intersection of the prime meridian with the equator lies on the positive x-axis. In this coordinate system, write the position vector of a place having latitude 30°N and longitude 45°E. Radius of the Earth is R.

Accepted Answer

Figure shows the coordinate system and point A.
∠AOB=θ (latitude) =30∘∠BOC=ϕ (longitude) =45∘.​
Resolving OA along the z-axis and OB, we get

Question 6

In a Cartesian coordinate system S, unit vectors along x and y axes are i^ and j^​. A new coordinate system S1​ is generated by rotating system S by an angle of 37∘ in the anticlockwise sense as shown in the figure.(i) Unit vectors in system S1​ are i^1​ and j^​1​. Express i^1​ and j^​1​ in terms of i^ and j^​.(ii) A vector in system S is expressed as r=2i^+2j^​. Express the vector in system S1​ using i^1​ and j^​1​.[ Take sin37∘≃0.6]

Accepted Answer

(i) i^1​=(1⋅cos37∘)i^+(1⋅sin37∘)j^​
⇒i^1​=0.8i^+0.6j^​ (a)
and j^​1​=−(1⋅sin37∘)i^+(1⋅cos37∘)j^​
⇒j^​1​=−0.6i^+0.8j^​ (b)

Question 7

Write vectors a,b,c and d using unit vectors i^,j^​. Magnitudes are as indicated. [tan37∘=43​]

Accepted Answer

Applying component of the vector 
a=52​i^+52​j^​ 
b=−63​i^+6j^​
c=−6i^−8j^​ 
d=3i^−4j^​

Question 8

Plot the vector a=2i^−2j^​ and calculate its magnitude.

Accepted Answer

Magnitude = 22+22​=22​

Question 9

Find the component of b in the direction of a
a=i^+j^​b=2i^+3j^​−k^​

Accepted Answer

Component of b in the direction of a = ∣a∣a⋅b​Hence, 2​5​

Question 10

F1​,F2​ and F3​ are three forces, represented as shown. Each of them is of equal magnitude. P,Q,R and S represent various directions

Accepted Answer

Break F1​ along x−y direction then apply addition and subtraction of vector. 
(c,d)

Question 11

For the situation shown below, which of the following results are correct?

Accepted Answer

Let force be F so resultant is in direction​4i^+3j^​+(5cos37∘i^+5sin37∘j^​)+F=ki^⇒4i^+3j^​+4i^+3j^​+F=ki^⇒8i^+6j^​+F=ki^⇒F=(k−8)i^−6j^​⇒F=(k−8)2+(6)2​⇒Fmin​=6N​

Question 12

Four vectors (A,B,C,D) all have the same magnitude and lie in a plane. The angle between adjacent vectors is 45∘ as shown. Which of the following equation(s) is/are correct?

Accepted Answer

Apply component form of addition of vector.
Break the B and D along the A and C
(a,b,d)

Question 13

Two forces are acting on a body as shown in the figure below. If the resultant of two forces lie along the y-axis,(a) find the magnitude of F2​.(b) what should be the direction of a third force F3​ so that the body stays under equilibrium?

Accepted Answer

F1​sin300=F2​cos370Solving equation F2​=425​(b) along positive Y axis

Question 14

If A=2i^+j^​ and B=i^−j^​, sketch vectors graphically and find the component of A along B and perpendicular to B

Accepted Answer

​21​(i^−j^​); Compo = ntofAperpendicar →B;23​(i^+j^​)A=2i^+j^​,B=i^−j^​​Component of A along B=BA⋅B​.B=Acosθ.B2​2−1​⋅2​(i^−j^​)​=21​(i^−j^​)Component of A perpendicular to B​=A−( compon ent of A along B)=2i^+j^​−21​(i^−j^​)=23​i^+23​j^​=23​(i^+j^​)​

Question 15

Three vectors as shown in the figure have magnitudes ∣a∣=3,∣b∣=4 and ∣c∣=10.(i) find the x and y components of these vectors.(ii) Find the numbers p and q such that c=pa+ qb

Accepted Answer

ax​=3,ay​=0;bx​=23​,by​=2;cx​=−5,cy​=53​;p=−20/3,q=53​/2]

Question 16

(a) Find a unit vector
in the xy plane at an angle of 45∘ with the x-axis.
(b) Resolve a vector R=2i^−3j^​ into two perpendicular components-one in a direction making 45∘ with +x-axis and other in a direction perpendicular to it. Express these two components using unit vectors in Cartesian coordinate system.

Accepted Answer

(a) Let unit vector be A which make angle 450 with x-axisA=2​1​i^+2​1​j^​(b) R make an angle with x-axis is θMagnitude of ∣R∣=1​3tanθ=2−3​cosθ=1​32​sinθ=1​3−3​Now, In new co-ordinate one component is Rcos(450+θ)=2​6−1​andother component is Rsin(450+θ)=2​6−1​

Question 17

A particle of 5 kg is momentarily at rest at x=0 at t=0. It is acted upon by two forces F1​ and F2​⋅F1​=70j^​ N. The direction and magnitude of F2​ are unknown. The particle experiences a constant acceleration, a, in the direction shown. Neglect gravity.(a) Find the missing force F2​.(b) What third force, F3​, is required to make the acceleration of the particle zero? Either give magnitude and direction of F3​ or its components.

Accepted Answer

F1​+F2​=m⋅a

(70j^​+F2​i^+F2​j^​)=5(10xcos53i^+10sin53j^​)

=50(10×53​i^+10×24​j^​)

By comparing two sides. We get

F2​i^=30i^⇒F2​j^​=−30j^​

∴F2​=30i^−30j^​

∣F2​∣=(30)2+(30)2​=302​N

At t=10s

v=u+at=10(6i^+8j^​)=60i^+80j^​

Acceleration will be zero if a third force makes resultant force =0

∴F+F2​+F+(3)=0

or F3​=−(F1​​+F2​)=−(70j^​+30i^−30j^​)=30i^−40j^​

Question 18

A force 250 N is to be resolved into components along line A−A′ and B−B′ oriented as shown in figure. If the resolved component along B B' equals 125 N. What would be the value of angle α ? Also, calculate the corresponding value of the force component along A-A'.

Accepted Answer

Along AA′250cosαAlong BB′250cos(450+α)=125cos(450+α)=250125​=21​(450+α)=600α=150

Filters

Filters

class

subject

topic

concept

difficulty

Solutions for component of vector Important Questions

Found the solution, but did not understand the concept?

Ask my first question