Free Body Diagrams
Trending Questions
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Column IColumn IIi. Force of friction is zero ina. Fig. (i)ii. Force of friction is 2.5 N inb. Fig. (ii)iii. Acceleration of the block is zero inc. Fig. (iiii)iv. Normal force is not equal to 2g in d. Fig. (iv)
- i−b, ii−a, iii−a, b, c, iv−c
- i−b, ii−c, d, iii−d, iv−a, d
- i−a, c, ii−b, d, iii−a, b, c, d, iv−c, d
- i−d, ii−a, b, iii−d, iv−c
- 62.5 N
- 72.5 N
- 82.5 N
- 92.5 N
- W2
- W
- W√3
- W2√3
is movable. If W1=W2=100 N, what is the angle (in degrees) AP2P1 when the system is in equilibrium?
The pulleys are massless and frictionless
Match the block diagrams of block of mass, m, with their corresponding free body diagram.
(Bodies are at rest with respect to ground)
P - (i); Q - (iii); R - (iv); S - (ii)
P - (ii); Q - (ii); R - (iv); S - (ii)
P - (i); Q - (iii); R - (iv); S - (iv)
None of these
- 400 N
- 374 N
- 418 N
- 448 N
- 1
- 2
- 3
- 4
- 4√3 and 20√3
- 4√3 and 20√3
- 4√3 and 20√3
- 4√3 and 20√30
- √3 m/s
- 2 m/s
- 2√3 m/s
- 3 m/s
Three blocks of masses 2 kg, 3 kg and 5 kg are connected to each other with light string and are then placed on a frictionless surface as shown in the figure. The system is pulled by a force F = 10 N, , then tension T1=
1N
5N
8N
10N
is movable. If W1=W2=100 N, what is the angle (in degrees) AP2P1 when the system is in equilibrium?
The pulleys are massless and frictionless
- Newton's third law of motion
- Newton's second law of motion
- Newton's first law of motion
- Newton's law of gravitation
Three masses of 15 kg. 10 kg and 5 kg are suspended vertically as shown in the fig. If the string attached to the support breaks and the system falls freely, what will be the tension in the string between 10 kg and 5 kg masses. Take g=10ms−2 . It is assumed that the string remains tight during the motion
300 N
250 N
50 N
Zero
- 400 N
- 800 N
- 300 N
- 500 N
- Tension
- Frictional force
- Weight of body
- Both (a) and (b)
- 3
- 4
- 5
- 6
(Take g=10 m/s2)
- 250 N
- Zero
- 500√3 N
- 500 N
- 1
- 2
- 3
- 4
- N will not have any component along inclined plane
- Block will not move
- Block will move in a direction perpendicular to inclined plane
- mg can not be resolved along the inclined plane
- mMgnm+M
- mMgnmM
- mg
- nmg
Statement II: From Newton’s Third Law, the force exerted by Block A on Block B is equal in magnitude to force exerted by block B on A.
- Both Statements I and II are true and Statement II is correct explanation of Statement I
- Both Statements I and II are true but Statement II is not correct explanation of Statement I
- Statement I is true and statement II is false
- Statement I is false and statement II is true
- 0∘
- 30∘
- 45∘
- 60∘
- 325 N
- 175 N
- 250 N
- 750 N
- 2
- 3
- 4
- 5
- mMgnm+M
- mMgnmM
- mg
- nmg
- 2g(1+μ1−μ)
- g(1+μ1−μ)
- g(1−μ1+μ)
- 2g(1−μ1+μ)
- 325 N
- 175 N
- 250 N
- 750 N
- 0∘
- 30∘
- 45∘
- 60∘
- 8 N
- 9 N
- 12 N
- Zero