What forces are acting on a block suspended by a rope over a pulley?

AI Thread Summary
The discussion centers on analyzing the forces acting on a block attached to a rope over a pulley. A block with a mass of 4.40 kg is on a frictionless surface, and the tension in the rope is measured at 18.5 N. The primary forces acting on the block include tension in the horizontal direction and gravitational force in the vertical direction. Participants emphasize the importance of drawing free body diagrams to identify all forces involved, including gravity acting downward. The correct application of Newton's second law is crucial for determining the acceleration of the blocks in this system.
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Homework Statement



A light rope is attached to a block with mass 4.40 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 18.5 N

Homework Equations



f=ma
Newtons 2nd law

The Attempt at a Solution



T - m(g) = ma

18.5 - 4.4(9.8) = 4.4(a)

a = 5.59

This is wrong?

Help please!

thank you :)
 
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mybrohshi5 said:

Homework Statement



A light rope is attached to a block with mass 4.40 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 18.5 N

Homework Equations



f=ma
Newtons 2nd law

The Attempt at a Solution



T - m(g) = ma

18.5 - 4.4(9.8) = 4.4(a)

a = 5.59

This is wrong?

Help please!

thank you :)
I gather you are trying to find the acceleration of the blocks...when you draw a free body diagram of the block on the table, what are the forces acting on the block in the X direction? Then use F_x_net = ma
 
i had to draw 2 free body diagrams for the first 2 parts of the assignment for the 4.4 kg mass and the unknown mass but i got them both wrong... :(

but the forces in the x direction acting on the block of mass 4.4 kg would be the tension and that is it right?
 
mybrohshi5 said:
i had to draw 2 free body diagrams for the first 2 parts of the assignment for the 4.4 kg mass and the unknown mass but i got them both wrong... :(

but the forces in the x direction acting on the block of mass 4.4 kg would be the tension and that is it right?
Right. That's the only force acting in the x direction. Ther are also forces in the y direction, 2 of them in the y direction. Can you identify them and their directions?
When you draw free body diagrams, identify all the forces acting. The force of gravity is usually one of them, acting down toward the center of the earth. This is an 'action at a distance' force; almost all other forces in Physics 101 are contact forces...there must be a contact between 2 surfaces or particles or objects in order for there to be a force.
 
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