145996
Three concurrent forces of the same magnitude are in equilibrium. What is the angle between the forces? Also, name the triangle formed by the forces as sides
1 \(60^{\circ}\) equilateral triangle
2 \(120^{\circ}\) equilateral triangle
3 \(120^{\circ}, 30^{\circ}, 30^{\circ}\) an isosceles triangle
4 \(120^{\circ}\) an obtuse triangle
Explanation:
B According to Lami's theorem, when three forces acting at a point are in equilibrium then each force is proportional to the sine of the angle between the other two forces. If forces \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are in equilibrium then- As the forces are equal in magnitude, \(\alpha=\beta=\gamma\) We know, \(\alpha+\beta+\gamma=360^{\circ}\) \(3 \alpha =360^{\circ}\) \(\alpha =120^{\circ}\) Therefore the angle \(\alpha=120^{\circ}, \beta=120^{\circ}\) and \(\square=120^{\circ}\) between the forces form equilateral triangle.
Manipal UGET-2018
Laws of Motion
145997
Consider an object kept at the centre, in the \(\boldsymbol{X} \boldsymbol{Y}\)-plane, on which five coplanar forces act as shown in the figure. The resultant force on the object is
1 \(6.5 \mathrm{~N}, 330^{\circ}\)
2 \(6.5 \mathrm{~N}, 300^{\circ}\)
3 \(6.5 \mathrm{~N}, 30^{\circ}\)
4 \(5.7 \mathrm{~N}, 331^{\circ}\)
Explanation:
A After resolving all forces into \(\mathrm{x}\) and \(\mathrm{y}\) plane, we get, Total force in \(\mathrm{x}\) - direction \(\mathrm{F}_{\mathrm{x}}=15 \cos 60^{\circ}+19-\left(16 \cos 45^{\circ}+11 \cos 30^{\circ}\right)\) \(=15 \times \frac{1}{2}+19-\left(16 \times \frac{1}{\sqrt{2}}+11 \times \frac{\sqrt{3}}{2}\right)\) \(=26.5-(11.31+9.52)\) \(=26.5-20.83=5.67 \mathrm{~N}\) Total Force in y - direction \(f_{y}=\left(16 \sin 45^{\circ}+15 \sin 60^{\circ}\right)-(11 \sin 30+22)\) \(\left(16 \times \frac{1}{\sqrt{2}}+15 \times \frac{\sqrt{3}}{2}\right)-\left(\frac{11}{2}+22\right)\) \(=(11.31+12.99)-(27.5)=-3.2 \mathrm{~N}\) Now, Resultant force \(\mathrm{F}=\sqrt{\mathrm{F}_{\mathrm{x}}^{2}+\mathrm{F}_{\mathrm{y}}^{2}}\) \(\mathrm{~F}=\sqrt{(5.67)^{2}+(-3.2)^{2}}=6.5 \mathrm{~N}\) Direction of Resultant force \(\theta=\tan ^{-1}\left(\frac{\mathrm{F}_{\mathrm{y}}}{\mathrm{F}_{\mathrm{x}}}\right)=\tan ^{-1}\left(\frac{-3.2}{5.67}\right)=330^{\circ}\)
TS- EAMCET-04.05.2018
Laws of Motion
145999
A mass of \(6 \mathrm{~kg}\) is suspended by a rope of length \(2 \mathrm{~m}\) from a ceiling. A force of \(50 \mathrm{~N}\) in the horizontal direction is applied at the mid- point \(P\) of the rope. The angle made by the rope with the vertical, in equilibrium is
1 \(50^{\circ}\)
2 \(60^{\circ}\)
3 \(30^{\circ}\)
4 \(40^{\circ}\)
5 \(45^{\circ}\)
Explanation:
D The given system is in equilibrium From figure, \(T_{1} \cos \theta=60 \mathrm{~N}\) \(\mathrm{T}_{1} \sin \theta=50 \mathrm{~N}\) Dividing equation (ii) by (i) we get - \(\frac{\mathrm{T}_{1} \sin \theta}{\mathrm{T}_{1} \cos \theta} =\frac{50}{60}\) \(\tan \theta =\frac{5}{6}\) \(\theta =\tan ^{-1}(5 / 6)=40^{\circ}\)
Kerala CEE 2007
Laws of Motion
146000
The sum of the magnitudes of two forces acting at a point is \(18 \mathrm{~N}\) and the magnitude of their resultant is \(12 \mathrm{~N}\). If the resultant is at \(90^{\circ}\) with the smaller force, the magnitude of the forces in \(\mathbf{N}\) are
1 6,12
2 11,7
3 5,13
4 14,4
5 10,8
Explanation:
C Let smaller force be \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) be the greater force then, \(\mathrm{F}_{1}+\mathrm{F}_{2}=18 \mathrm{~N}\) Resultant \(\mathrm{R}\) of the forces is \(90^{\circ}\) to \(\mathrm{AB}\) (i.e. \(\mathrm{F}_{1}\) ), \(\text { or } (12)^{2}=\mathrm{F}_{2}^{2}-\mathrm{F}_{1}^{2}\) \(\text { or } 144=\left(\mathrm{F}_{2}-\mathrm{F}_{1}\right)\left(\mathrm{F}_{2}+\mathrm{F}_{1}\right)\) \(\therefore \left(\mathrm{F}_{2}-\mathrm{F}_{1}\right) \times 18=144\) \(\left(\mathrm{~F}_{2}-\mathrm{F}_{1}\right)=\frac{144}{18}=8\) Adding equation (i) and (ii), we get \(\mathrm{F}_{1}+\mathrm{F}_{2}+\mathrm{F}_{2}-\mathrm{F}_{1}=18+8\) \(2 \mathrm{~F}_{2}=26\) \(\mathrm{~F}_{2}=13 \mathrm{~N}\) Putting the value of \(\mathrm{F}_{2}\) in equation (i) we get - \(\mathrm{F}_{1}+13=18\) \(\mathrm{F}_{1}=5 \mathrm{~N}\)
Kerala CEE 2007
Laws of Motion
146001
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then :
1 Pattern A is more sturdy
2 Pattern B is more sturdy
3 Pattern C is more sturdy
4 all will have same sturdiness
Explanation:
A Due to same weight of street light in all three cases torque will be same. If \(\mathrm{T}\) is tension and \(l\) be perpendicular distance of cable from the axis then \(\tau =\mathrm{T} \times l\) \(\mathrm{~T} \times l=\text { constant } \quad(\tau \text { is constant })\) \(\mathrm{T} \propto \frac{1}{l}\) Hence, tension will be least for largest cable. So pattern A is more sturdy.
145996
Three concurrent forces of the same magnitude are in equilibrium. What is the angle between the forces? Also, name the triangle formed by the forces as sides
1 \(60^{\circ}\) equilateral triangle
2 \(120^{\circ}\) equilateral triangle
3 \(120^{\circ}, 30^{\circ}, 30^{\circ}\) an isosceles triangle
4 \(120^{\circ}\) an obtuse triangle
Explanation:
B According to Lami's theorem, when three forces acting at a point are in equilibrium then each force is proportional to the sine of the angle between the other two forces. If forces \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are in equilibrium then- As the forces are equal in magnitude, \(\alpha=\beta=\gamma\) We know, \(\alpha+\beta+\gamma=360^{\circ}\) \(3 \alpha =360^{\circ}\) \(\alpha =120^{\circ}\) Therefore the angle \(\alpha=120^{\circ}, \beta=120^{\circ}\) and \(\square=120^{\circ}\) between the forces form equilateral triangle.
Manipal UGET-2018
Laws of Motion
145997
Consider an object kept at the centre, in the \(\boldsymbol{X} \boldsymbol{Y}\)-plane, on which five coplanar forces act as shown in the figure. The resultant force on the object is
1 \(6.5 \mathrm{~N}, 330^{\circ}\)
2 \(6.5 \mathrm{~N}, 300^{\circ}\)
3 \(6.5 \mathrm{~N}, 30^{\circ}\)
4 \(5.7 \mathrm{~N}, 331^{\circ}\)
Explanation:
A After resolving all forces into \(\mathrm{x}\) and \(\mathrm{y}\) plane, we get, Total force in \(\mathrm{x}\) - direction \(\mathrm{F}_{\mathrm{x}}=15 \cos 60^{\circ}+19-\left(16 \cos 45^{\circ}+11 \cos 30^{\circ}\right)\) \(=15 \times \frac{1}{2}+19-\left(16 \times \frac{1}{\sqrt{2}}+11 \times \frac{\sqrt{3}}{2}\right)\) \(=26.5-(11.31+9.52)\) \(=26.5-20.83=5.67 \mathrm{~N}\) Total Force in y - direction \(f_{y}=\left(16 \sin 45^{\circ}+15 \sin 60^{\circ}\right)-(11 \sin 30+22)\) \(\left(16 \times \frac{1}{\sqrt{2}}+15 \times \frac{\sqrt{3}}{2}\right)-\left(\frac{11}{2}+22\right)\) \(=(11.31+12.99)-(27.5)=-3.2 \mathrm{~N}\) Now, Resultant force \(\mathrm{F}=\sqrt{\mathrm{F}_{\mathrm{x}}^{2}+\mathrm{F}_{\mathrm{y}}^{2}}\) \(\mathrm{~F}=\sqrt{(5.67)^{2}+(-3.2)^{2}}=6.5 \mathrm{~N}\) Direction of Resultant force \(\theta=\tan ^{-1}\left(\frac{\mathrm{F}_{\mathrm{y}}}{\mathrm{F}_{\mathrm{x}}}\right)=\tan ^{-1}\left(\frac{-3.2}{5.67}\right)=330^{\circ}\)
TS- EAMCET-04.05.2018
Laws of Motion
145999
A mass of \(6 \mathrm{~kg}\) is suspended by a rope of length \(2 \mathrm{~m}\) from a ceiling. A force of \(50 \mathrm{~N}\) in the horizontal direction is applied at the mid- point \(P\) of the rope. The angle made by the rope with the vertical, in equilibrium is
1 \(50^{\circ}\)
2 \(60^{\circ}\)
3 \(30^{\circ}\)
4 \(40^{\circ}\)
5 \(45^{\circ}\)
Explanation:
D The given system is in equilibrium From figure, \(T_{1} \cos \theta=60 \mathrm{~N}\) \(\mathrm{T}_{1} \sin \theta=50 \mathrm{~N}\) Dividing equation (ii) by (i) we get - \(\frac{\mathrm{T}_{1} \sin \theta}{\mathrm{T}_{1} \cos \theta} =\frac{50}{60}\) \(\tan \theta =\frac{5}{6}\) \(\theta =\tan ^{-1}(5 / 6)=40^{\circ}\)
Kerala CEE 2007
Laws of Motion
146000
The sum of the magnitudes of two forces acting at a point is \(18 \mathrm{~N}\) and the magnitude of their resultant is \(12 \mathrm{~N}\). If the resultant is at \(90^{\circ}\) with the smaller force, the magnitude of the forces in \(\mathbf{N}\) are
1 6,12
2 11,7
3 5,13
4 14,4
5 10,8
Explanation:
C Let smaller force be \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) be the greater force then, \(\mathrm{F}_{1}+\mathrm{F}_{2}=18 \mathrm{~N}\) Resultant \(\mathrm{R}\) of the forces is \(90^{\circ}\) to \(\mathrm{AB}\) (i.e. \(\mathrm{F}_{1}\) ), \(\text { or } (12)^{2}=\mathrm{F}_{2}^{2}-\mathrm{F}_{1}^{2}\) \(\text { or } 144=\left(\mathrm{F}_{2}-\mathrm{F}_{1}\right)\left(\mathrm{F}_{2}+\mathrm{F}_{1}\right)\) \(\therefore \left(\mathrm{F}_{2}-\mathrm{F}_{1}\right) \times 18=144\) \(\left(\mathrm{~F}_{2}-\mathrm{F}_{1}\right)=\frac{144}{18}=8\) Adding equation (i) and (ii), we get \(\mathrm{F}_{1}+\mathrm{F}_{2}+\mathrm{F}_{2}-\mathrm{F}_{1}=18+8\) \(2 \mathrm{~F}_{2}=26\) \(\mathrm{~F}_{2}=13 \mathrm{~N}\) Putting the value of \(\mathrm{F}_{2}\) in equation (i) we get - \(\mathrm{F}_{1}+13=18\) \(\mathrm{F}_{1}=5 \mathrm{~N}\)
Kerala CEE 2007
Laws of Motion
146001
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then :
1 Pattern A is more sturdy
2 Pattern B is more sturdy
3 Pattern C is more sturdy
4 all will have same sturdiness
Explanation:
A Due to same weight of street light in all three cases torque will be same. If \(\mathrm{T}\) is tension and \(l\) be perpendicular distance of cable from the axis then \(\tau =\mathrm{T} \times l\) \(\mathrm{~T} \times l=\text { constant } \quad(\tau \text { is constant })\) \(\mathrm{T} \propto \frac{1}{l}\) Hence, tension will be least for largest cable. So pattern A is more sturdy.
145996
Three concurrent forces of the same magnitude are in equilibrium. What is the angle between the forces? Also, name the triangle formed by the forces as sides
1 \(60^{\circ}\) equilateral triangle
2 \(120^{\circ}\) equilateral triangle
3 \(120^{\circ}, 30^{\circ}, 30^{\circ}\) an isosceles triangle
4 \(120^{\circ}\) an obtuse triangle
Explanation:
B According to Lami's theorem, when three forces acting at a point are in equilibrium then each force is proportional to the sine of the angle between the other two forces. If forces \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are in equilibrium then- As the forces are equal in magnitude, \(\alpha=\beta=\gamma\) We know, \(\alpha+\beta+\gamma=360^{\circ}\) \(3 \alpha =360^{\circ}\) \(\alpha =120^{\circ}\) Therefore the angle \(\alpha=120^{\circ}, \beta=120^{\circ}\) and \(\square=120^{\circ}\) between the forces form equilateral triangle.
Manipal UGET-2018
Laws of Motion
145997
Consider an object kept at the centre, in the \(\boldsymbol{X} \boldsymbol{Y}\)-plane, on which five coplanar forces act as shown in the figure. The resultant force on the object is
1 \(6.5 \mathrm{~N}, 330^{\circ}\)
2 \(6.5 \mathrm{~N}, 300^{\circ}\)
3 \(6.5 \mathrm{~N}, 30^{\circ}\)
4 \(5.7 \mathrm{~N}, 331^{\circ}\)
Explanation:
A After resolving all forces into \(\mathrm{x}\) and \(\mathrm{y}\) plane, we get, Total force in \(\mathrm{x}\) - direction \(\mathrm{F}_{\mathrm{x}}=15 \cos 60^{\circ}+19-\left(16 \cos 45^{\circ}+11 \cos 30^{\circ}\right)\) \(=15 \times \frac{1}{2}+19-\left(16 \times \frac{1}{\sqrt{2}}+11 \times \frac{\sqrt{3}}{2}\right)\) \(=26.5-(11.31+9.52)\) \(=26.5-20.83=5.67 \mathrm{~N}\) Total Force in y - direction \(f_{y}=\left(16 \sin 45^{\circ}+15 \sin 60^{\circ}\right)-(11 \sin 30+22)\) \(\left(16 \times \frac{1}{\sqrt{2}}+15 \times \frac{\sqrt{3}}{2}\right)-\left(\frac{11}{2}+22\right)\) \(=(11.31+12.99)-(27.5)=-3.2 \mathrm{~N}\) Now, Resultant force \(\mathrm{F}=\sqrt{\mathrm{F}_{\mathrm{x}}^{2}+\mathrm{F}_{\mathrm{y}}^{2}}\) \(\mathrm{~F}=\sqrt{(5.67)^{2}+(-3.2)^{2}}=6.5 \mathrm{~N}\) Direction of Resultant force \(\theta=\tan ^{-1}\left(\frac{\mathrm{F}_{\mathrm{y}}}{\mathrm{F}_{\mathrm{x}}}\right)=\tan ^{-1}\left(\frac{-3.2}{5.67}\right)=330^{\circ}\)
TS- EAMCET-04.05.2018
Laws of Motion
145999
A mass of \(6 \mathrm{~kg}\) is suspended by a rope of length \(2 \mathrm{~m}\) from a ceiling. A force of \(50 \mathrm{~N}\) in the horizontal direction is applied at the mid- point \(P\) of the rope. The angle made by the rope with the vertical, in equilibrium is
1 \(50^{\circ}\)
2 \(60^{\circ}\)
3 \(30^{\circ}\)
4 \(40^{\circ}\)
5 \(45^{\circ}\)
Explanation:
D The given system is in equilibrium From figure, \(T_{1} \cos \theta=60 \mathrm{~N}\) \(\mathrm{T}_{1} \sin \theta=50 \mathrm{~N}\) Dividing equation (ii) by (i) we get - \(\frac{\mathrm{T}_{1} \sin \theta}{\mathrm{T}_{1} \cos \theta} =\frac{50}{60}\) \(\tan \theta =\frac{5}{6}\) \(\theta =\tan ^{-1}(5 / 6)=40^{\circ}\)
Kerala CEE 2007
Laws of Motion
146000
The sum of the magnitudes of two forces acting at a point is \(18 \mathrm{~N}\) and the magnitude of their resultant is \(12 \mathrm{~N}\). If the resultant is at \(90^{\circ}\) with the smaller force, the magnitude of the forces in \(\mathbf{N}\) are
1 6,12
2 11,7
3 5,13
4 14,4
5 10,8
Explanation:
C Let smaller force be \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) be the greater force then, \(\mathrm{F}_{1}+\mathrm{F}_{2}=18 \mathrm{~N}\) Resultant \(\mathrm{R}\) of the forces is \(90^{\circ}\) to \(\mathrm{AB}\) (i.e. \(\mathrm{F}_{1}\) ), \(\text { or } (12)^{2}=\mathrm{F}_{2}^{2}-\mathrm{F}_{1}^{2}\) \(\text { or } 144=\left(\mathrm{F}_{2}-\mathrm{F}_{1}\right)\left(\mathrm{F}_{2}+\mathrm{F}_{1}\right)\) \(\therefore \left(\mathrm{F}_{2}-\mathrm{F}_{1}\right) \times 18=144\) \(\left(\mathrm{~F}_{2}-\mathrm{F}_{1}\right)=\frac{144}{18}=8\) Adding equation (i) and (ii), we get \(\mathrm{F}_{1}+\mathrm{F}_{2}+\mathrm{F}_{2}-\mathrm{F}_{1}=18+8\) \(2 \mathrm{~F}_{2}=26\) \(\mathrm{~F}_{2}=13 \mathrm{~N}\) Putting the value of \(\mathrm{F}_{2}\) in equation (i) we get - \(\mathrm{F}_{1}+13=18\) \(\mathrm{F}_{1}=5 \mathrm{~N}\)
Kerala CEE 2007
Laws of Motion
146001
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then :
1 Pattern A is more sturdy
2 Pattern B is more sturdy
3 Pattern C is more sturdy
4 all will have same sturdiness
Explanation:
A Due to same weight of street light in all three cases torque will be same. If \(\mathrm{T}\) is tension and \(l\) be perpendicular distance of cable from the axis then \(\tau =\mathrm{T} \times l\) \(\mathrm{~T} \times l=\text { constant } \quad(\tau \text { is constant })\) \(\mathrm{T} \propto \frac{1}{l}\) Hence, tension will be least for largest cable. So pattern A is more sturdy.
145996
Three concurrent forces of the same magnitude are in equilibrium. What is the angle between the forces? Also, name the triangle formed by the forces as sides
1 \(60^{\circ}\) equilateral triangle
2 \(120^{\circ}\) equilateral triangle
3 \(120^{\circ}, 30^{\circ}, 30^{\circ}\) an isosceles triangle
4 \(120^{\circ}\) an obtuse triangle
Explanation:
B According to Lami's theorem, when three forces acting at a point are in equilibrium then each force is proportional to the sine of the angle between the other two forces. If forces \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are in equilibrium then- As the forces are equal in magnitude, \(\alpha=\beta=\gamma\) We know, \(\alpha+\beta+\gamma=360^{\circ}\) \(3 \alpha =360^{\circ}\) \(\alpha =120^{\circ}\) Therefore the angle \(\alpha=120^{\circ}, \beta=120^{\circ}\) and \(\square=120^{\circ}\) between the forces form equilateral triangle.
Manipal UGET-2018
Laws of Motion
145997
Consider an object kept at the centre, in the \(\boldsymbol{X} \boldsymbol{Y}\)-plane, on which five coplanar forces act as shown in the figure. The resultant force on the object is
1 \(6.5 \mathrm{~N}, 330^{\circ}\)
2 \(6.5 \mathrm{~N}, 300^{\circ}\)
3 \(6.5 \mathrm{~N}, 30^{\circ}\)
4 \(5.7 \mathrm{~N}, 331^{\circ}\)
Explanation:
A After resolving all forces into \(\mathrm{x}\) and \(\mathrm{y}\) plane, we get, Total force in \(\mathrm{x}\) - direction \(\mathrm{F}_{\mathrm{x}}=15 \cos 60^{\circ}+19-\left(16 \cos 45^{\circ}+11 \cos 30^{\circ}\right)\) \(=15 \times \frac{1}{2}+19-\left(16 \times \frac{1}{\sqrt{2}}+11 \times \frac{\sqrt{3}}{2}\right)\) \(=26.5-(11.31+9.52)\) \(=26.5-20.83=5.67 \mathrm{~N}\) Total Force in y - direction \(f_{y}=\left(16 \sin 45^{\circ}+15 \sin 60^{\circ}\right)-(11 \sin 30+22)\) \(\left(16 \times \frac{1}{\sqrt{2}}+15 \times \frac{\sqrt{3}}{2}\right)-\left(\frac{11}{2}+22\right)\) \(=(11.31+12.99)-(27.5)=-3.2 \mathrm{~N}\) Now, Resultant force \(\mathrm{F}=\sqrt{\mathrm{F}_{\mathrm{x}}^{2}+\mathrm{F}_{\mathrm{y}}^{2}}\) \(\mathrm{~F}=\sqrt{(5.67)^{2}+(-3.2)^{2}}=6.5 \mathrm{~N}\) Direction of Resultant force \(\theta=\tan ^{-1}\left(\frac{\mathrm{F}_{\mathrm{y}}}{\mathrm{F}_{\mathrm{x}}}\right)=\tan ^{-1}\left(\frac{-3.2}{5.67}\right)=330^{\circ}\)
TS- EAMCET-04.05.2018
Laws of Motion
145999
A mass of \(6 \mathrm{~kg}\) is suspended by a rope of length \(2 \mathrm{~m}\) from a ceiling. A force of \(50 \mathrm{~N}\) in the horizontal direction is applied at the mid- point \(P\) of the rope. The angle made by the rope with the vertical, in equilibrium is
1 \(50^{\circ}\)
2 \(60^{\circ}\)
3 \(30^{\circ}\)
4 \(40^{\circ}\)
5 \(45^{\circ}\)
Explanation:
D The given system is in equilibrium From figure, \(T_{1} \cos \theta=60 \mathrm{~N}\) \(\mathrm{T}_{1} \sin \theta=50 \mathrm{~N}\) Dividing equation (ii) by (i) we get - \(\frac{\mathrm{T}_{1} \sin \theta}{\mathrm{T}_{1} \cos \theta} =\frac{50}{60}\) \(\tan \theta =\frac{5}{6}\) \(\theta =\tan ^{-1}(5 / 6)=40^{\circ}\)
Kerala CEE 2007
Laws of Motion
146000
The sum of the magnitudes of two forces acting at a point is \(18 \mathrm{~N}\) and the magnitude of their resultant is \(12 \mathrm{~N}\). If the resultant is at \(90^{\circ}\) with the smaller force, the magnitude of the forces in \(\mathbf{N}\) are
1 6,12
2 11,7
3 5,13
4 14,4
5 10,8
Explanation:
C Let smaller force be \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) be the greater force then, \(\mathrm{F}_{1}+\mathrm{F}_{2}=18 \mathrm{~N}\) Resultant \(\mathrm{R}\) of the forces is \(90^{\circ}\) to \(\mathrm{AB}\) (i.e. \(\mathrm{F}_{1}\) ), \(\text { or } (12)^{2}=\mathrm{F}_{2}^{2}-\mathrm{F}_{1}^{2}\) \(\text { or } 144=\left(\mathrm{F}_{2}-\mathrm{F}_{1}\right)\left(\mathrm{F}_{2}+\mathrm{F}_{1}\right)\) \(\therefore \left(\mathrm{F}_{2}-\mathrm{F}_{1}\right) \times 18=144\) \(\left(\mathrm{~F}_{2}-\mathrm{F}_{1}\right)=\frac{144}{18}=8\) Adding equation (i) and (ii), we get \(\mathrm{F}_{1}+\mathrm{F}_{2}+\mathrm{F}_{2}-\mathrm{F}_{1}=18+8\) \(2 \mathrm{~F}_{2}=26\) \(\mathrm{~F}_{2}=13 \mathrm{~N}\) Putting the value of \(\mathrm{F}_{2}\) in equation (i) we get - \(\mathrm{F}_{1}+13=18\) \(\mathrm{F}_{1}=5 \mathrm{~N}\)
Kerala CEE 2007
Laws of Motion
146001
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then :
1 Pattern A is more sturdy
2 Pattern B is more sturdy
3 Pattern C is more sturdy
4 all will have same sturdiness
Explanation:
A Due to same weight of street light in all three cases torque will be same. If \(\mathrm{T}\) is tension and \(l\) be perpendicular distance of cable from the axis then \(\tau =\mathrm{T} \times l\) \(\mathrm{~T} \times l=\text { constant } \quad(\tau \text { is constant })\) \(\mathrm{T} \propto \frac{1}{l}\) Hence, tension will be least for largest cable. So pattern A is more sturdy.
145996
Three concurrent forces of the same magnitude are in equilibrium. What is the angle between the forces? Also, name the triangle formed by the forces as sides
1 \(60^{\circ}\) equilateral triangle
2 \(120^{\circ}\) equilateral triangle
3 \(120^{\circ}, 30^{\circ}, 30^{\circ}\) an isosceles triangle
4 \(120^{\circ}\) an obtuse triangle
Explanation:
B According to Lami's theorem, when three forces acting at a point are in equilibrium then each force is proportional to the sine of the angle between the other two forces. If forces \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are in equilibrium then- As the forces are equal in magnitude, \(\alpha=\beta=\gamma\) We know, \(\alpha+\beta+\gamma=360^{\circ}\) \(3 \alpha =360^{\circ}\) \(\alpha =120^{\circ}\) Therefore the angle \(\alpha=120^{\circ}, \beta=120^{\circ}\) and \(\square=120^{\circ}\) between the forces form equilateral triangle.
Manipal UGET-2018
Laws of Motion
145997
Consider an object kept at the centre, in the \(\boldsymbol{X} \boldsymbol{Y}\)-plane, on which five coplanar forces act as shown in the figure. The resultant force on the object is
1 \(6.5 \mathrm{~N}, 330^{\circ}\)
2 \(6.5 \mathrm{~N}, 300^{\circ}\)
3 \(6.5 \mathrm{~N}, 30^{\circ}\)
4 \(5.7 \mathrm{~N}, 331^{\circ}\)
Explanation:
A After resolving all forces into \(\mathrm{x}\) and \(\mathrm{y}\) plane, we get, Total force in \(\mathrm{x}\) - direction \(\mathrm{F}_{\mathrm{x}}=15 \cos 60^{\circ}+19-\left(16 \cos 45^{\circ}+11 \cos 30^{\circ}\right)\) \(=15 \times \frac{1}{2}+19-\left(16 \times \frac{1}{\sqrt{2}}+11 \times \frac{\sqrt{3}}{2}\right)\) \(=26.5-(11.31+9.52)\) \(=26.5-20.83=5.67 \mathrm{~N}\) Total Force in y - direction \(f_{y}=\left(16 \sin 45^{\circ}+15 \sin 60^{\circ}\right)-(11 \sin 30+22)\) \(\left(16 \times \frac{1}{\sqrt{2}}+15 \times \frac{\sqrt{3}}{2}\right)-\left(\frac{11}{2}+22\right)\) \(=(11.31+12.99)-(27.5)=-3.2 \mathrm{~N}\) Now, Resultant force \(\mathrm{F}=\sqrt{\mathrm{F}_{\mathrm{x}}^{2}+\mathrm{F}_{\mathrm{y}}^{2}}\) \(\mathrm{~F}=\sqrt{(5.67)^{2}+(-3.2)^{2}}=6.5 \mathrm{~N}\) Direction of Resultant force \(\theta=\tan ^{-1}\left(\frac{\mathrm{F}_{\mathrm{y}}}{\mathrm{F}_{\mathrm{x}}}\right)=\tan ^{-1}\left(\frac{-3.2}{5.67}\right)=330^{\circ}\)
TS- EAMCET-04.05.2018
Laws of Motion
145999
A mass of \(6 \mathrm{~kg}\) is suspended by a rope of length \(2 \mathrm{~m}\) from a ceiling. A force of \(50 \mathrm{~N}\) in the horizontal direction is applied at the mid- point \(P\) of the rope. The angle made by the rope with the vertical, in equilibrium is
1 \(50^{\circ}\)
2 \(60^{\circ}\)
3 \(30^{\circ}\)
4 \(40^{\circ}\)
5 \(45^{\circ}\)
Explanation:
D The given system is in equilibrium From figure, \(T_{1} \cos \theta=60 \mathrm{~N}\) \(\mathrm{T}_{1} \sin \theta=50 \mathrm{~N}\) Dividing equation (ii) by (i) we get - \(\frac{\mathrm{T}_{1} \sin \theta}{\mathrm{T}_{1} \cos \theta} =\frac{50}{60}\) \(\tan \theta =\frac{5}{6}\) \(\theta =\tan ^{-1}(5 / 6)=40^{\circ}\)
Kerala CEE 2007
Laws of Motion
146000
The sum of the magnitudes of two forces acting at a point is \(18 \mathrm{~N}\) and the magnitude of their resultant is \(12 \mathrm{~N}\). If the resultant is at \(90^{\circ}\) with the smaller force, the magnitude of the forces in \(\mathbf{N}\) are
1 6,12
2 11,7
3 5,13
4 14,4
5 10,8
Explanation:
C Let smaller force be \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) be the greater force then, \(\mathrm{F}_{1}+\mathrm{F}_{2}=18 \mathrm{~N}\) Resultant \(\mathrm{R}\) of the forces is \(90^{\circ}\) to \(\mathrm{AB}\) (i.e. \(\mathrm{F}_{1}\) ), \(\text { or } (12)^{2}=\mathrm{F}_{2}^{2}-\mathrm{F}_{1}^{2}\) \(\text { or } 144=\left(\mathrm{F}_{2}-\mathrm{F}_{1}\right)\left(\mathrm{F}_{2}+\mathrm{F}_{1}\right)\) \(\therefore \left(\mathrm{F}_{2}-\mathrm{F}_{1}\right) \times 18=144\) \(\left(\mathrm{~F}_{2}-\mathrm{F}_{1}\right)=\frac{144}{18}=8\) Adding equation (i) and (ii), we get \(\mathrm{F}_{1}+\mathrm{F}_{2}+\mathrm{F}_{2}-\mathrm{F}_{1}=18+8\) \(2 \mathrm{~F}_{2}=26\) \(\mathrm{~F}_{2}=13 \mathrm{~N}\) Putting the value of \(\mathrm{F}_{2}\) in equation (i) we get - \(\mathrm{F}_{1}+13=18\) \(\mathrm{F}_{1}=5 \mathrm{~N}\)
Kerala CEE 2007
Laws of Motion
146001
If a street light of mass \(M\) is suspended from the end of a uniform rod of length \(L\) in different possible patterns as shown in figure, then :
1 Pattern A is more sturdy
2 Pattern B is more sturdy
3 Pattern C is more sturdy
4 all will have same sturdiness
Explanation:
A Due to same weight of street light in all three cases torque will be same. If \(\mathrm{T}\) is tension and \(l\) be perpendicular distance of cable from the axis then \(\tau =\mathrm{T} \times l\) \(\mathrm{~T} \times l=\text { constant } \quad(\tau \text { is constant })\) \(\mathrm{T} \propto \frac{1}{l}\) Hence, tension will be least for largest cable. So pattern A is more sturdy.
