Horizontal force due to tension from vertical wieght

In summary, the problem involves three blocks, C (4 kg), B (2 kg), and A (1 kg) on a frictionless horizontal surface with a rope and pulley system. The goal is to find the horizontal force F needed to push block C so that block A rises with an upward acceleration of 3 m/s^2. Using a free body diagram and the equation F=ma, the solution is found to be F=38.4 N. However, the actual answer is 59.8 N, which may be due to a mistake in the mass of blocks A and B.
  • #1
fatalphysics
1
0

Homework Statement



Block C (m= 4 kg) sits on a frictionless horizontal surface. Block B of (m
= 2kg) sits on top of block C, and is attached to a rope that runs over the massless
pulley as shown in the figure. Block A (m=1 kg) hangs vertically from the rope. With
what horizontal force F must you push Block C so that block A rises with an upward
acceleration of a = 3m/s2? All surfaces are frictionless.

Diagram:
question7pic.png

Homework Equations



F=ma

The Attempt at a Solution



Using free body diagrams came up with:

F=mac
T=2ac
T/2=ac

F=maa
T-mg=(3)(1)
T=3+(9.8)(1)
T=12.8 N

(12.8)/2=ac
ac=6.4 m/s2

F=(mc+ma)ac
F=(6)(6.4)
F=38.4 N

Actual answer = 59.8N
 
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  • #2
fatalphysics said:

Homework Statement



Block C (m= 4 kg) sits on a frictionless horizontal surface. Block B of (m
= 2kg) sits on top of block C, and is attached to a rope that runs over the massless
pulley as shown in the figure. Block A (m=1 kg) hangs vertically from the rope. With
what horizontal force F must you push Block C so that block A rises with an upward
acceleration of a = 3m/s2? All surfaces are frictionless.

Diagram:
View attachment 40143

Homework Equations



F=ma

The Attempt at a Solution



Using free body diagrams came up with:

F=mac
T=2ac
T/2=ac

F=maa
T-mg=(3)(1)
T=3+(9.8)(1)
T=12.8 N

(12.8)/2=ac
ac=6.4 m/s2

F=(mc+ma)ac
F=(6)(6.4)
F=38.4 N

Actual answer = 59.8N

It looks like you are mixing up the mass of A and B. A is 1 ; B is 2 not the other way.
 

What is horizontal force due to tension from vertical weight?

The horizontal force due to tension from vertical weight is the force that acts horizontally on an object due to the tension created by its own weight pulling downwards.

How is horizontal force due to tension from vertical weight calculated?

The calculation of horizontal force due to tension from vertical weight involves using the formula F = mg tanθ, where F is the horizontal force, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle between the horizontal and the direction of the tension force.

What factors affect the magnitude of horizontal force due to tension from vertical weight?

The magnitude of horizontal force due to tension from vertical weight is affected by the mass of the object, the strength of the tension force, the angle of the tension force, and the acceleration due to gravity.

How does horizontal force due to tension from vertical weight impact an object's motion?

The horizontal force due to tension from vertical weight can cause an object to accelerate in the horizontal direction, change its direction of motion, or remain in a state of equilibrium depending on the balance of forces acting on the object.

What are some real-world applications of horizontal force due to tension from vertical weight?

Horizontal force due to tension from vertical weight is commonly seen in situations where cables or ropes are used to support objects, such as in elevators, cranes, and suspension bridges. It is also a key concept in understanding the physics of pulleys and other simple machines.

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