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Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards?
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Two blocks of masses M 1 and M 2 are connected with a string...
Analysis:
To determine the minimum mass M2 required for the block M1 to slide upwards, we need to consider the forces acting on the system and the conditions for equilibrium.
Forces Acting on the System:
1. Weight (mg1) acting vertically downwards on M1.
2. Normal force (N) acting perpendicular to the incline.
3. Friction force (F) acting parallel to the incline and opposing the motion.
4. Tension force (T) acting on the string connecting M1 and M2.
5. Weight (mg2) acting vertically downwards on M2.
Conditions for Equilibrium:
For M1 to slide upwards, we need to ensure that the net force acting on M1 is directed upwards. This can be achieved by satisfying the following conditions:
1. The component of weight of M1 along the incline (mg1sinθ) must be greater than the friction force (F).
2. The component of weight of M2 along the incline (mg2sinθ) must be less than the tension force (T).
Calculation:
1. Net Force on M1:
- Along the incline: F_net = mg1sinθ - F
- Vertically: F_net_vert = mg1cosθ - N
- Since M1 is in equilibrium, F_net = 0 and F_net_vert = 0.
- From F_net_vert equation, N = mg1cosθ.
2. Friction Force:
- F = μN = μmg1cosθ.
3. Net Force on M2:
- Along the incline: F_net = mg2sinθ - T
- Vertically: F_net_vert = mg2cosθ - N
- Since M2 is in equilibrium, F_net = 0 and F_net_vert = 0.
- From F_net_vert equation, N = mg2cosθ.
4. Setting up the Equations:
- For M1: mg1sinθ - μmg1cosθ = 0. (Equation 1)
- For M2: mg2sinθ - T = 0. (Equation 2)
- From Equation 1, we can solve for μmg1cosθ.
- Substituting the value of N in Equation 2, we get: mg2sinθ - μmg1cosθ = 0.
- Rearranging, μmg1cosθ = mg2sinθ.
5. Minimum Mass M2:
- To ensure M1 slides upwards, we need to minimize the tension force T.
- For this, we need to maximize the value of μmg1cosθ.
- From the equation μmg1cosθ = mg2sinθ, we can see that increasing M2 will increase the value of mg2sinθ.
- Therefore, the minimum mass M2 required for M1 to slide upwards is M2 = μM1.
Conclusion:
The minimum mass M2 required for the block M1 to slide upwards is M2 = μM1.
To determine the minimum mass M2 required for the block M1 to slide upwards, we need to consider the forces acting on the system and the conditions for equilibrium.
Forces Acting on the System:
1. Weight (mg1) acting vertically downwards on M1.
2. Normal force (N) acting perpendicular to the incline.
3. Friction force (F) acting parallel to the incline and opposing the motion.
4. Tension force (T) acting on the string connecting M1 and M2.
5. Weight (mg2) acting vertically downwards on M2.
Conditions for Equilibrium:
For M1 to slide upwards, we need to ensure that the net force acting on M1 is directed upwards. This can be achieved by satisfying the following conditions:
1. The component of weight of M1 along the incline (mg1sinθ) must be greater than the friction force (F).
2. The component of weight of M2 along the incline (mg2sinθ) must be less than the tension force (T).
Calculation:
1. Net Force on M1:
- Along the incline: F_net = mg1sinθ - F
- Vertically: F_net_vert = mg1cosθ - N
- Since M1 is in equilibrium, F_net = 0 and F_net_vert = 0.
- From F_net_vert equation, N = mg1cosθ.
2. Friction Force:
- F = μN = μmg1cosθ.
3. Net Force on M2:
- Along the incline: F_net = mg2sinθ - T
- Vertically: F_net_vert = mg2cosθ - N
- Since M2 is in equilibrium, F_net = 0 and F_net_vert = 0.
- From F_net_vert equation, N = mg2cosθ.
4. Setting up the Equations:
- For M1: mg1sinθ - μmg1cosθ = 0. (Equation 1)
- For M2: mg2sinθ - T = 0. (Equation 2)
- From Equation 1, we can solve for μmg1cosθ.
- Substituting the value of N in Equation 2, we get: mg2sinθ - μmg1cosθ = 0.
- Rearranging, μmg1cosθ = mg2sinθ.
5. Minimum Mass M2:
- To ensure M1 slides upwards, we need to minimize the tension force T.
- For this, we need to maximize the value of μmg1cosθ.
- From the equation μmg1cosθ = mg2sinθ, we can see that increasing M2 will increase the value of mg2sinθ.
- Therefore, the minimum mass M2 required for M1 to slide upwards is M2 = μM1.
Conclusion:
The minimum mass M2 required for the block M1 to slide upwards is M2 = μM1.
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Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards?
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Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards?.
Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards?.
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Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards?, a detailed solution for Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards? has been provided alongside types of Two blocks of masses M 1 and M 2 are connected with a string which passes over a smooth pulley. The mass M 1 is placed on a rough incline plane as shown in Fig. The coefficient of friction between the block and the inclined planes is μ. What should be the minimum mass M 2 so that the block M 1 slides upwards? theory, EduRev gives you an
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