363268
Which of the following groups of forces could be in equilibrium?
1 \(3N, 4N, 5N\)
2 \(4N, 5N, 10N\)
3 \(30N, 40N, 80N\)
4 \(1N, 3N, 5N\)
Explanation:
For equilibrium of forces, the resultant of two (smaller) forces should be equal and opposite to third one. This is only possible with option (1)
PHXI05:LAWS OF MOTION
363269
A mass of 5 \(kg\) is suspended by a rope of length 2 \(m\) from a ceiling. A force of 50 \(N\) in the horizontal direction is applied at the mid-point 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 \(45^\circ \)
Explanation:
The free body diagram of the rope is shown in the figure. \(\tan \theta = \frac{{50}}{{50}} = 1\) \( \Rightarrow \quad \tan \theta = \tan 45^\circ \) \( \Rightarrow \quad \theta = 45^\circ \)
PHXI05:LAWS OF MOTION
363270
Three forces acting on a body are shown in the figure. To have the resultant force only along the \(y\)- direction, the magnitude of the minimum additional force needed is
1 \(0.5\,N\)
2 \(1.5\,N\)
3 \(\frac{{\sqrt 3 }}{4}N\)
4 \(\sqrt 3 N\)
Explanation:
\( \Rightarrow \) Net force along \(x\)-axis \( = + (1\cos 60^\circ + 2\sin 30^\circ ) - 4\sin 30^\circ \) \( = \left( {\frac{1}{2} + 2 \times \frac{1}{2}} \right) - 4 \times \frac{1}{2} = \frac{1}{2}\) Net force along \(y\)-axis \( = 4\cos 30^\circ + 1\sin 60^\circ - 2\cos 30^\circ \) \( \Rightarrow 4 \times \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2} - 2 \times \frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3 }}{2}\) To have, resultant only in \(y\)-axis we must have \(\frac{1}{2}N = 0.5\) force towards \( + x\)-axis, so that it can compensate the net force of \( - x\) axis.
PHXI05:LAWS OF MOTION
363271
Two smooth spheres each of radius 5 \(cm \) and weight \(W\) rest one on the other inside a fixed smooth cylinder of radius 8 \(cm\). The reactions between the sphere and the vertical side of the cylinder are:
The FBD’s of \(A\) and \(B\) are: For the horizontal equlibrium of the system \({N_{AW}} = {N_{BW}}\) For sphere A \({N_{AB}}\sin \theta + W = {N_{AG}}\) \({N_{AB}}\cos \theta = {N_{AW}}\) For sphere B \({N_{AB}}\cos \theta = {N_{BW}}\) \({N_{AB}}\sin \theta = W\) \(\cos \theta = \frac{6}{{10}} \Rightarrow \theta = 53^\circ \) From the above eq’s \({N_{AW}} = {N_{BW}} = \frac{{3W}}{4}\)
PHXI05:LAWS OF MOTION
363272
A block of mass 4 \(kg\) is at rest on a rough inclined plane as show in the figure. The magnitude of net force exerted by the surface on the block will be
1 \(26\,N\)
2 \(19.5\,N\)
3 \(10\,N\)
4 \(40\,N\)
Explanation:
Net force applied by block on the inclined plane is equal to the weight of the body
363268
Which of the following groups of forces could be in equilibrium?
1 \(3N, 4N, 5N\)
2 \(4N, 5N, 10N\)
3 \(30N, 40N, 80N\)
4 \(1N, 3N, 5N\)
Explanation:
For equilibrium of forces, the resultant of two (smaller) forces should be equal and opposite to third one. This is only possible with option (1)
PHXI05:LAWS OF MOTION
363269
A mass of 5 \(kg\) is suspended by a rope of length 2 \(m\) from a ceiling. A force of 50 \(N\) in the horizontal direction is applied at the mid-point 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 \(45^\circ \)
Explanation:
The free body diagram of the rope is shown in the figure. \(\tan \theta = \frac{{50}}{{50}} = 1\) \( \Rightarrow \quad \tan \theta = \tan 45^\circ \) \( \Rightarrow \quad \theta = 45^\circ \)
PHXI05:LAWS OF MOTION
363270
Three forces acting on a body are shown in the figure. To have the resultant force only along the \(y\)- direction, the magnitude of the minimum additional force needed is
1 \(0.5\,N\)
2 \(1.5\,N\)
3 \(\frac{{\sqrt 3 }}{4}N\)
4 \(\sqrt 3 N\)
Explanation:
\( \Rightarrow \) Net force along \(x\)-axis \( = + (1\cos 60^\circ + 2\sin 30^\circ ) - 4\sin 30^\circ \) \( = \left( {\frac{1}{2} + 2 \times \frac{1}{2}} \right) - 4 \times \frac{1}{2} = \frac{1}{2}\) Net force along \(y\)-axis \( = 4\cos 30^\circ + 1\sin 60^\circ - 2\cos 30^\circ \) \( \Rightarrow 4 \times \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2} - 2 \times \frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3 }}{2}\) To have, resultant only in \(y\)-axis we must have \(\frac{1}{2}N = 0.5\) force towards \( + x\)-axis, so that it can compensate the net force of \( - x\) axis.
PHXI05:LAWS OF MOTION
363271
Two smooth spheres each of radius 5 \(cm \) and weight \(W\) rest one on the other inside a fixed smooth cylinder of radius 8 \(cm\). The reactions between the sphere and the vertical side of the cylinder are:
The FBD’s of \(A\) and \(B\) are: For the horizontal equlibrium of the system \({N_{AW}} = {N_{BW}}\) For sphere A \({N_{AB}}\sin \theta + W = {N_{AG}}\) \({N_{AB}}\cos \theta = {N_{AW}}\) For sphere B \({N_{AB}}\cos \theta = {N_{BW}}\) \({N_{AB}}\sin \theta = W\) \(\cos \theta = \frac{6}{{10}} \Rightarrow \theta = 53^\circ \) From the above eq’s \({N_{AW}} = {N_{BW}} = \frac{{3W}}{4}\)
PHXI05:LAWS OF MOTION
363272
A block of mass 4 \(kg\) is at rest on a rough inclined plane as show in the figure. The magnitude of net force exerted by the surface on the block will be
1 \(26\,N\)
2 \(19.5\,N\)
3 \(10\,N\)
4 \(40\,N\)
Explanation:
Net force applied by block on the inclined plane is equal to the weight of the body
363268
Which of the following groups of forces could be in equilibrium?
1 \(3N, 4N, 5N\)
2 \(4N, 5N, 10N\)
3 \(30N, 40N, 80N\)
4 \(1N, 3N, 5N\)
Explanation:
For equilibrium of forces, the resultant of two (smaller) forces should be equal and opposite to third one. This is only possible with option (1)
PHXI05:LAWS OF MOTION
363269
A mass of 5 \(kg\) is suspended by a rope of length 2 \(m\) from a ceiling. A force of 50 \(N\) in the horizontal direction is applied at the mid-point 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 \(45^\circ \)
Explanation:
The free body diagram of the rope is shown in the figure. \(\tan \theta = \frac{{50}}{{50}} = 1\) \( \Rightarrow \quad \tan \theta = \tan 45^\circ \) \( \Rightarrow \quad \theta = 45^\circ \)
PHXI05:LAWS OF MOTION
363270
Three forces acting on a body are shown in the figure. To have the resultant force only along the \(y\)- direction, the magnitude of the minimum additional force needed is
1 \(0.5\,N\)
2 \(1.5\,N\)
3 \(\frac{{\sqrt 3 }}{4}N\)
4 \(\sqrt 3 N\)
Explanation:
\( \Rightarrow \) Net force along \(x\)-axis \( = + (1\cos 60^\circ + 2\sin 30^\circ ) - 4\sin 30^\circ \) \( = \left( {\frac{1}{2} + 2 \times \frac{1}{2}} \right) - 4 \times \frac{1}{2} = \frac{1}{2}\) Net force along \(y\)-axis \( = 4\cos 30^\circ + 1\sin 60^\circ - 2\cos 30^\circ \) \( \Rightarrow 4 \times \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2} - 2 \times \frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3 }}{2}\) To have, resultant only in \(y\)-axis we must have \(\frac{1}{2}N = 0.5\) force towards \( + x\)-axis, so that it can compensate the net force of \( - x\) axis.
PHXI05:LAWS OF MOTION
363271
Two smooth spheres each of radius 5 \(cm \) and weight \(W\) rest one on the other inside a fixed smooth cylinder of radius 8 \(cm\). The reactions between the sphere and the vertical side of the cylinder are:
The FBD’s of \(A\) and \(B\) are: For the horizontal equlibrium of the system \({N_{AW}} = {N_{BW}}\) For sphere A \({N_{AB}}\sin \theta + W = {N_{AG}}\) \({N_{AB}}\cos \theta = {N_{AW}}\) For sphere B \({N_{AB}}\cos \theta = {N_{BW}}\) \({N_{AB}}\sin \theta = W\) \(\cos \theta = \frac{6}{{10}} \Rightarrow \theta = 53^\circ \) From the above eq’s \({N_{AW}} = {N_{BW}} = \frac{{3W}}{4}\)
PHXI05:LAWS OF MOTION
363272
A block of mass 4 \(kg\) is at rest on a rough inclined plane as show in the figure. The magnitude of net force exerted by the surface on the block will be
1 \(26\,N\)
2 \(19.5\,N\)
3 \(10\,N\)
4 \(40\,N\)
Explanation:
Net force applied by block on the inclined plane is equal to the weight of the body
363268
Which of the following groups of forces could be in equilibrium?
1 \(3N, 4N, 5N\)
2 \(4N, 5N, 10N\)
3 \(30N, 40N, 80N\)
4 \(1N, 3N, 5N\)
Explanation:
For equilibrium of forces, the resultant of two (smaller) forces should be equal and opposite to third one. This is only possible with option (1)
PHXI05:LAWS OF MOTION
363269
A mass of 5 \(kg\) is suspended by a rope of length 2 \(m\) from a ceiling. A force of 50 \(N\) in the horizontal direction is applied at the mid-point 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 \(45^\circ \)
Explanation:
The free body diagram of the rope is shown in the figure. \(\tan \theta = \frac{{50}}{{50}} = 1\) \( \Rightarrow \quad \tan \theta = \tan 45^\circ \) \( \Rightarrow \quad \theta = 45^\circ \)
PHXI05:LAWS OF MOTION
363270
Three forces acting on a body are shown in the figure. To have the resultant force only along the \(y\)- direction, the magnitude of the minimum additional force needed is
1 \(0.5\,N\)
2 \(1.5\,N\)
3 \(\frac{{\sqrt 3 }}{4}N\)
4 \(\sqrt 3 N\)
Explanation:
\( \Rightarrow \) Net force along \(x\)-axis \( = + (1\cos 60^\circ + 2\sin 30^\circ ) - 4\sin 30^\circ \) \( = \left( {\frac{1}{2} + 2 \times \frac{1}{2}} \right) - 4 \times \frac{1}{2} = \frac{1}{2}\) Net force along \(y\)-axis \( = 4\cos 30^\circ + 1\sin 60^\circ - 2\cos 30^\circ \) \( \Rightarrow 4 \times \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2} - 2 \times \frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3 }}{2}\) To have, resultant only in \(y\)-axis we must have \(\frac{1}{2}N = 0.5\) force towards \( + x\)-axis, so that it can compensate the net force of \( - x\) axis.
PHXI05:LAWS OF MOTION
363271
Two smooth spheres each of radius 5 \(cm \) and weight \(W\) rest one on the other inside a fixed smooth cylinder of radius 8 \(cm\). The reactions between the sphere and the vertical side of the cylinder are:
The FBD’s of \(A\) and \(B\) are: For the horizontal equlibrium of the system \({N_{AW}} = {N_{BW}}\) For sphere A \({N_{AB}}\sin \theta + W = {N_{AG}}\) \({N_{AB}}\cos \theta = {N_{AW}}\) For sphere B \({N_{AB}}\cos \theta = {N_{BW}}\) \({N_{AB}}\sin \theta = W\) \(\cos \theta = \frac{6}{{10}} \Rightarrow \theta = 53^\circ \) From the above eq’s \({N_{AW}} = {N_{BW}} = \frac{{3W}}{4}\)
PHXI05:LAWS OF MOTION
363272
A block of mass 4 \(kg\) is at rest on a rough inclined plane as show in the figure. The magnitude of net force exerted by the surface on the block will be
1 \(26\,N\)
2 \(19.5\,N\)
3 \(10\,N\)
4 \(40\,N\)
Explanation:
Net force applied by block on the inclined plane is equal to the weight of the body
363268
Which of the following groups of forces could be in equilibrium?
1 \(3N, 4N, 5N\)
2 \(4N, 5N, 10N\)
3 \(30N, 40N, 80N\)
4 \(1N, 3N, 5N\)
Explanation:
For equilibrium of forces, the resultant of two (smaller) forces should be equal and opposite to third one. This is only possible with option (1)
PHXI05:LAWS OF MOTION
363269
A mass of 5 \(kg\) is suspended by a rope of length 2 \(m\) from a ceiling. A force of 50 \(N\) in the horizontal direction is applied at the mid-point 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 \(45^\circ \)
Explanation:
The free body diagram of the rope is shown in the figure. \(\tan \theta = \frac{{50}}{{50}} = 1\) \( \Rightarrow \quad \tan \theta = \tan 45^\circ \) \( \Rightarrow \quad \theta = 45^\circ \)
PHXI05:LAWS OF MOTION
363270
Three forces acting on a body are shown in the figure. To have the resultant force only along the \(y\)- direction, the magnitude of the minimum additional force needed is
1 \(0.5\,N\)
2 \(1.5\,N\)
3 \(\frac{{\sqrt 3 }}{4}N\)
4 \(\sqrt 3 N\)
Explanation:
\( \Rightarrow \) Net force along \(x\)-axis \( = + (1\cos 60^\circ + 2\sin 30^\circ ) - 4\sin 30^\circ \) \( = \left( {\frac{1}{2} + 2 \times \frac{1}{2}} \right) - 4 \times \frac{1}{2} = \frac{1}{2}\) Net force along \(y\)-axis \( = 4\cos 30^\circ + 1\sin 60^\circ - 2\cos 30^\circ \) \( \Rightarrow 4 \times \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2} - 2 \times \frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3 }}{2}\) To have, resultant only in \(y\)-axis we must have \(\frac{1}{2}N = 0.5\) force towards \( + x\)-axis, so that it can compensate the net force of \( - x\) axis.
PHXI05:LAWS OF MOTION
363271
Two smooth spheres each of radius 5 \(cm \) and weight \(W\) rest one on the other inside a fixed smooth cylinder of radius 8 \(cm\). The reactions between the sphere and the vertical side of the cylinder are:
The FBD’s of \(A\) and \(B\) are: For the horizontal equlibrium of the system \({N_{AW}} = {N_{BW}}\) For sphere A \({N_{AB}}\sin \theta + W = {N_{AG}}\) \({N_{AB}}\cos \theta = {N_{AW}}\) For sphere B \({N_{AB}}\cos \theta = {N_{BW}}\) \({N_{AB}}\sin \theta = W\) \(\cos \theta = \frac{6}{{10}} \Rightarrow \theta = 53^\circ \) From the above eq’s \({N_{AW}} = {N_{BW}} = \frac{{3W}}{4}\)
PHXI05:LAWS OF MOTION
363272
A block of mass 4 \(kg\) is at rest on a rough inclined plane as show in the figure. The magnitude of net force exerted by the surface on the block will be
1 \(26\,N\)
2 \(19.5\,N\)
3 \(10\,N\)
4 \(40\,N\)
Explanation:
Net force applied by block on the inclined plane is equal to the weight of the body
