A force problem that seems easy but hard


by cy19861126
Tags: force
cy19861126
cy19861126 is offline
#1
Nov2-06, 07:24 PM
P: 69
A 2.00-kg aluminum block and a 6.00-kg copper block are connected by a light string over a frictionless pulley. The two blocks are allowed to move on a fixed steel block wedge (of angle 30.0) as shown in Figure P4.53. Making use of Table 4.2, determine (a) the acceleration of the two blocks and (b) the tension in the string.

My work so far:
For the aluminum block - I got T = 2a, which I cannot solve
For the copper block - I got Tx-58.8sin(30) = 6a - this is for x direction
and
Ny-58.8cos(30) = 6a - this is for y direction

58.8 is the weight force, which comes from 6*9.8 = 58.8N
T = tension force
N = Normal force

All the equations won't get me to solve for acceleration for either blocks

Thanks~~

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Fig 4.53.JPG  
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QuantumCrash
QuantumCrash is offline
#2
Nov3-06, 07:45 AM
P: 135
You don't need an x for the direction of T for the copper block since you are resolving it along the surface of the plane. Once you get this, its semple simultaneous equations.
cy19861126
cy19861126 is offline
#3
Nov3-06, 06:22 PM
P: 69
Quote Quote by QuantumCrash
You don't need an x for the direction of T for the copper block since you are resolving it along the surface of the plane. Once you get this, its semple simultaneous equations.
I see what you are saying, but for the copper block, I defined the x axis to be the the ramp. Therefore, the coordinate diagram is a slanted one, not a standard one. As a result, you need Tx

turdferguson
turdferguson is offline
#4
Nov3-06, 10:47 PM
P: 312

A force problem that seems easy but hard


Neither block is accelerating in the y as you have defined it. Tension is purely in the x. You then get 2 equations with 2 unknowns because the tensions and accelerations are equal
QuantumCrash
QuantumCrash is offline
#5
Nov4-06, 10:35 PM
P: 135
If I get what you mean, you should not do that becaus you would need to take into account the normal force.


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