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.
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Laws of Motion
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.
