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A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)
- a)200 N
- b)100 N
- c)140 N
- d)70 N
Correct answer is option 'B'. Can you explain this answer?
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A mass of 10 kg is suspended vertically by a rope from the roof. When ...
at equation
F = 100 N
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A mass of 10 kg is suspended vertically by a rope from the roof. When ...
Degrees from the vertical. The tension in the rope is 100 N. What is the magnitude and direction of the force applied horizontally?
To solve this problem, we can analyze the forces acting on the mass. There are two forces acting on the mass: the tension force in the rope and the force applied horizontally.
First, let's find the vertical component of the tension force. Since the rope deviates at an angle of 45 degrees from the vertical, the vertical component of the tension force can be found using trigonometry:
Vertical component of tension force = Tension force * sin(angle)
Vertical component of tension force = 100 N * sin(45 degrees)
Vertical component of tension force ≈ 100 N * 0.707
Vertical component of tension force ≈ 70.7 N
Now, let's analyze the forces acting vertically. The mass is in equilibrium, so the weight force must be balanced by the vertical component of the tension force:
Weight force = Vertical component of tension force
Mass * gravitational acceleration = Vertical component of tension force
10 kg * 9.8 m/s^2 = 70.7 N
98 N = 70.7 N
Since the weight force and the vertical component of the tension force are equal, the vertical forces are balanced.
Next, let's analyze the forces acting horizontally. The horizontal component of the tension force must be balanced by the force applied horizontally:
Horizontal component of tension force = Force applied horizontally
Horizontal component of tension force = 100 N * cos(angle)
Horizontal component of tension force = 100 N * cos(45 degrees)
Horizontal component of tension force ≈ 100 N * 0.707
Horizontal component of tension force ≈ 70.7 N
Since the horizontal component of the tension force and the force applied horizontally are equal, the horizontal forces are balanced.
Therefore, the magnitude of the force applied horizontally is 70.7 N, and the direction of the force is horizontal.
To solve this problem, we can analyze the forces acting on the mass. There are two forces acting on the mass: the tension force in the rope and the force applied horizontally.
First, let's find the vertical component of the tension force. Since the rope deviates at an angle of 45 degrees from the vertical, the vertical component of the tension force can be found using trigonometry:
Vertical component of tension force = Tension force * sin(angle)
Vertical component of tension force = 100 N * sin(45 degrees)
Vertical component of tension force ≈ 100 N * 0.707
Vertical component of tension force ≈ 70.7 N
Now, let's analyze the forces acting vertically. The mass is in equilibrium, so the weight force must be balanced by the vertical component of the tension force:
Weight force = Vertical component of tension force
Mass * gravitational acceleration = Vertical component of tension force
10 kg * 9.8 m/s^2 = 70.7 N
98 N = 70.7 N
Since the weight force and the vertical component of the tension force are equal, the vertical forces are balanced.
Next, let's analyze the forces acting horizontally. The horizontal component of the tension force must be balanced by the force applied horizontally:
Horizontal component of tension force = Force applied horizontally
Horizontal component of tension force = 100 N * cos(angle)
Horizontal component of tension force = 100 N * cos(45 degrees)
Horizontal component of tension force ≈ 100 N * 0.707
Horizontal component of tension force ≈ 70.7 N
Since the horizontal component of the tension force and the force applied horizontally are equal, the horizontal forces are balanced.
Therefore, the magnitude of the force applied horizontally is 70.7 N, and the direction of the force is horizontal.
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A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)a)200 Nb)100 Nc)140 Nd)70 NCorrect answer is option 'B'. Can you explain this answer?
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A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)a)200 Nb)100 Nc)140 Nd)70 NCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)a)200 Nb)100 Nc)140 Nd)70 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)a)200 Nb)100 Nc)140 Nd)70 NCorrect answer is option 'B'. Can you explain this answer?.
A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)a)200 Nb)100 Nc)140 Nd)70 NCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)a)200 Nb)100 Nc)140 Nd)70 NCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms –2)a)200 Nb)100 Nc)140 Nd)70 NCorrect answer is option 'B'. Can you explain this answer?.
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