Mathematics - Statics
Exam Duration: 45 Mins Total Questions : 30
The resultant of two forces P and Q is R. If Q is doubled,then R is doubled. If direction of Q is reversed, R is doubled again, then \({ P }^{ 2 }:{ Q }^{ 2 }:{ R }^{ 2 }\)is:
- (a)
2:3:2
- (b)
1:2:3
- (c)
2:3:1
- (d)
3:1:1
If the resultant of two forces of magnitudes P and 2P is perpendicular to P, then the angle between the forces, is
- (a)
\(\frac { 2\Pi }{ 3 } \)
- (b)
\(\frac { 3\Pi }{ 4 } \)
- (c)
\(\frac { 4\Pi }{ 5 } \)
- (d)
\(\frac { 5\Pi }{ 6 } \)
Two forces P and Q acting at a point are such that if one of these is reversed, the direction of the resultant is turned through a right angle, then
- (a)
P=2Q
- (b)
2P=Q
- (c)
2P=3Q
- (d)
P=Q
Two forces P+Q and P-Q make an angle 2\(\alpha \) with each other and their resultant makes an angles \(\theta \) with the bisector of the angle between them, then
- (a)
\(P\quad \tan { \theta } =Q\quad \tan { \alpha } \)
- (b)
\(P\quad \tan { \alpha } =Q\quad \tan { \theta } \)
- (c)
\(P\quad \tan { (\alpha +\theta ) } =Q\quad \tan { (\alpha } -\theta )\)
- (d)
None of these
The greatest resultant of two forces is P and least resultant is Q. when these forces act an angle \(\theta\), then the resultant ,is
- (a)
\(\sqrt { ({ P }^{ 2 }{ sin }^{ 2 }(\frac { \theta }{ 2 } )+{ Q }^{ 2 }{ cos }^{ 2 }(\frac { \theta }{ 2 } ) } \)
- (b)
\(\sqrt { ({ P }^{ 2 }{ +Q }^{ 2 } } \)
- (c)
\(\sqrt { { P }^{ 2\quad }{ cos }^{ 2 }(\frac { \theta }{ 2 } )+{ Q }^{ 2 }{ sin }^{ 2 }(\frac { \theta }{ 2 } ) } \)
- (d)
None of these
The resultant of two forces P and Q is \(\sqrt { 3 }\)Q and makes an angle \({ 30 }^{ \circ }\)with the direction of P. then
- (a)
P=Q or P=2Q
- (b)
P=Q or Q=2P
- (c)
P=2Q or Q=3P
- (d)
P=Q or P=4Q
The resultant R of two forces P and Q act at right angles to P. Then the angle between these forces, is
- (a)
\({ cos }^{ -1 }(\frac { P }{ Q } )\)
- (b)
\({ cos }^{ -1 }(-\frac { P }{ Q } )\)
- (c)
\({ sin }^{ -1 }(\frac { P }{ Q } )\)
- (d)
\({ sin }^{ -1 }(-\frac { P }{ Q } )\)
If the resultant of two forces each of a unit magnitude,acting at a point ,is of unit magnitude, then angle between these forces, is
- (a)
\({ 45 }^{ \circ }\)
- (b)
\({ 60 }^{ \circ }\)
- (c)
\({ 90 }^{ \circ }\)
- (d)
\({ 120 }^{ \circ }\)
The resultant of two equals forces acting on a particle is equal to three times the product of these forces. then angle between these forces is:
- (a)
\({ 30 }^{ \circ }\)
- (b)
\({ 45 }^{ \circ }\)
- (c)
\({ 60 }^{ \circ }\)
- (d)
\({ 90 }^{ \circ }\)
Three forces of magnitude 8N,5N and 4N acting at a point are in equilibrium, then the angle between smaller forces is:
- (a)
\({ cos }^{ -1 }(\frac { 21 }{ 40 } )\)
- (b)
\({ cos }^{ -1 }(\frac { 23 }{ 40 } )\)
- (c)
\({ cos }^{ -1 }(\frac { 20 }{ 39 } )\)
- (d)
\({ cos }^{ -1 }(\frac { 23 }{ 39 } )\)
The line of action of the resultant of two forces P and Q divides the angle between them in the ration 1:2. then the magnitude of their resultant is:
- (a)
\(\frac { { Q }^{ 2 }-{ P }^{ 2 } }{ Q } \)
- (b)
\(\frac { { P }^{ 2 }-{ Q }^{ 2 } }{ P } \)
- (c)
\(\frac { { P }^{ 2 }-{ Q }^{ 2 } }{ Q } \)
- (d)
\(\frac { { P }^{ 2 }+{ Q }^{ 2 } }{ { P }^{ 2 }-{ Q }^{ 2 } } \)
Three forces P,2P and 3P act along the sides AB,BC,CA of a given equilateral triangle ABC. then magnitude of the resultant, is.
- (a)
0
- (b)
\(\frac { \sqrt { 3 } }{ 2 } P\)
- (c)
\( \sqrt { 3 } P\)
- (d)
None of these
Three like parallel forcesP,Q and R act at the vertices of a triangle ABC.If the line of action of the resultant passes through the orthocentre
- (a)
\(\frac { P } {a }=\frac { Q } {b }=\frac { R } {c }\)
- (b)
\(\frac { P } {sin 2A }=\frac { Q } {sin2b }=\frac { R } {sin2c }\)
- (c)
\(\frac { P } {tanA }=\frac { Q } {tanB }=\frac { R } {tanC }\)
- (d)
None of these
Two like parallel forces 5N and 15N, act on a light rod at two points A and B respectively, 6m apart.The magnitude of resultant and the distance of its point of action from the point A, are respectively
- (a)
10N;4.5m
- (b)
20N;4.5m
- (c)
20N;1.5m
- (d)
10N;1.5m
P and Q are two like parallel forces.If P is moved parallel to itself through a distance x,then resultant of these forces moves through a distance
- (a)
\(\frac { Qx }{ P+Q } \)
- (b)
\(\frac { x }{ P+Q } \)
- (c)
(P+Q)x
- (d)
\(\frac { Px }{ P+Q } \)
Three like forces P,Q,R act at vertices of a triangle ABC.If their resultant passes through the circumcentre of \(\Delta ABC\),then
- (a)
\(\frac { P }{ a } =\frac { Q }{ b } =\frac { R }{ c } \)
- (b)
\(\frac { P }{ sin2A } =\frac { Q }{ sin2B } =\frac { R }{ sin2C } \)
- (c)
\(\frac { P }{ tan A } =\frac { Q }{ tan B } =\frac { R }{ tan C } \)
- (d)
None of these
A man can exert a force of 100N and pulls on a rope fastened to the top of a post, the length of the rope beiong twice the length of the post.The horizontal force applied to the middle of the post that will keep it fromfailing, is
- (a)
100N
- (b)
200N
- (c)
\(100\sqrt { 3}\)N
- (d)
200N
Three forces P,Q,R act along the sides BC,CA and AB of the triangle ABC.If their resultant passes through their centroid, then
- (a)
\(\frac { P }{ a } =\frac { Q }{ b } =\frac { R }{ c } =0\)
- (b)
P sec A+Q sec B+R sec C=0
- (c)
P cot A+Q cot B+R cot C=0
- (d)
P tan A+Q tan B+R tan C=0
Forces P and Q whose resultant is R act at point O. A transveral is drawn meeting lines of action of these forces at L,M and N respectively.then,\(\frac { p} {OL}+\frac { Q} { OM }\)equals
- (a)
\(\frac { R } { PL+OM}\)
- (b)
\(\frac { R } { ON }\)
- (c)
\(\frac { R } { OL-OM }\)
- (d)
None of these
Two forces acting at the point A (-3,0) and B(3,0) form a couple of moment 30 units.If AB is the arm of the couple then magnitude of each of these forces, is
- (a)
5 units
- (b)
6 units
- (c)
8 units
- (d)
10 units
ABCD is a rectangle such that AB=CD=a and BC=DA=b. forces P,P act along AD and CB and forces Q, Q act along AB and CD. Then perpendicular distance of force P and Q at A the resultant of forces P,Q at C, is:
- (a)
\(\frac { Qb-Pa }{ \sqrt { { P }^{ 2 }+{ Q }^{ 2 } } } \)
- (b)
\(\frac { Qb+Pa }{ \sqrt { { P }^{ 2 }+{ Q }^{ 2 } } } \)
- (c)
\(\frac { Pa-Qb }{ \sqrt { { P }^{ 2 }+{ Q }^{ 2 } } } \)
- (d)
None of these
Two forces P and Q acting parallel to the length and base of inclined plane respectively would each of them singly support weight W on the plane;then \(\frac { 1 }{ { P }^{ 2 } } -\frac { 1 }{ { Q }^{ 2 } } \)equals
- (a)
\(\frac { 2 }{ { W }^{ 2 } } \)
- (b)
\(\frac { 4 }{ { W }^{ 2 } } \)
- (c)
\(\frac { 1 }{ { W }^{ 2 } } \)
- (d)
None of these
A horizontal force F is applied to a small object P of mass m on a smooth plane inclined at an angle \(\theta\) to the horizontal. If F is just enough to keep theobject in equilibrium, then magnitude of F is
- (a)
\(mg\quad { cos }^{ 2 }\theta \)
- (b)
\(mg\quad { sin }^{ 2 }\theta \)
- (c)
\(mg { cos }\theta \)
- (d)
\(mg\quad { tan }\theta\)
A practicle is acted upon by three forces P,Q and R.It cannot be in equilibrium, if P:Q:R, is
- (a)
1:3:5
- (b)
3:5:7
- (c)
5:7:9
- (d)
7:9:11
A couple is of moment \(\overrightarrow { G } \) and the force forming the couple is \(\overrightarrow { P } \).If \(\overrightarrow { P } \) is turned a right angle the moment of the couple thus formed is \(\overrightarrow { H }\). If instead, the forces \(\overrightarrow { P } \). are turned through an angle \(\alpha\), then the moment of the couple is:
- (a)
\(\overrightarrow { H } \cos { \alpha } +\overrightarrow { G } \sin { \alpha } \)
- (b)
\(\overrightarrow { G } \cos { \alpha } +\overrightarrow { H } \sin { \alpha } \)
- (c)
\(\overrightarrow { H } \sin { \alpha } -\overrightarrow { G } \cos { \alpha } \)
- (d)
\(\overrightarrow { G } \sin { \alpha } +\overrightarrow { H } \cos { \alpha } \)
Two forces acting at the points A and B form a couple of moment of G.If their lines of action are turned through a right angle, these forces form a couple of moment H. Now, if both act at right angles to Ab,then the moment of the new couple is:
- (a)
\(\sqrt { { G }^{ 2 }+{ H }^{ 2 } } \)
- (b)
\( { { G }^{ 2 }+{ H }^{ 2 } } \)
- (c)
\(\frac { { G }^{ 2 } }{ { H }^{ 2 } } \)
- (d)
\( { G }^{ 2 } { H }^{ 2 } \)
If three forces acting on a rigid body are represented in magnitude, direction and line of action by the sides of a triangle, taken in order.If these forces form a couple whose moments is equal to K times the area of the triangle ,then K equals
- (a)
4
- (b)
3
- (c)
2
- (d)
1