- #1

- 261

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I have taken the direction of motion of each mass as the positive side.

So this is what I did,

For m=6kg

Fnety= Fg-T=6a

Fg-T=6a

For m= 5 Kg

Fnetx= T-Ff=5a

This is what i could do. I am unclear on how to use the angle..Any help?

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- Thread starter pinkyjoshi65
- Start date

- #1

- 261

- 0

I have taken the direction of motion of each mass as the positive side.

So this is what I did,

For m=6kg

Fnety= Fg-T=6a

Fg-T=6a

For m= 5 Kg

Fnetx= T-Ff=5a

This is what i could do. I am unclear on how to use the angle..Any help?

- #2

Doc Al

Mentor

- 45,135

- 1,433

What is Ff? (There's no friction.) Don't forget the component of gravity parallel to the incline. (That's where the angle comes in.)For m= 5 Kg

Fnetx= T-Ff=5a

- #3

- 261

- 0

uhh..so it will be something like this?

For m=5Kg

Fa=mgSinTheta and Fn=mgcosTheta?

For m=5Kg

Fa=mgSinTheta and Fn=mgcosTheta?

- #4

Doc Al

Mentor

- 45,135

- 1,433

- #5

- 261

- 0

so, mgsinTheta is the only force for m=5kg, we have to take into account?

- #6

Doc Al

Mentor

- 45,135

- 1,433

No, you still have the rope pulling it up the ramp.so, mgsinTheta is the only force for m=5kg, we have to take into account?

- #7

- 261

- 0

yea so the net force is T-mgsinTheta?

- #8

Doc Al

Mentor

- 45,135

- 1,433

Yes.yea so the net force is T-mgsinTheta?

- #9

- 261

- 0

one more question: a similar type

Three blocks (12 kg, 6.0 kg and 4.0 kg) are connected through a frictionless system. Find the tension in the two strings and the acceleration of the system.

So for the acceleration:

Fnetx of system=T=ma

T1+T2=6a

T1=6a-T2

Fy= 117.6-T1 (for m=12kg)

Fy= T2-39.2 (for m=4 kg)

Net Fy for the system= 78.4+T2-T1=ma

78.4+T2-6a+T2=16a

78.4+2T2= 22a

This is where i got stuck

Three blocks (12 kg, 6.0 kg and 4.0 kg) are connected through a frictionless system. Find the tension in the two strings and the acceleration of the system.

So for the acceleration:

Fnetx of system=T=ma

T1+T2=6a

T1=6a-T2

Fy= 117.6-T1 (for m=12kg)

Fy= T2-39.2 (for m=4 kg)

Net Fy for the system= 78.4+T2-T1=ma

78.4+T2-6a+T2=16a

78.4+2T2= 22a

This is where i got stuck

- #10

Doc Al

Mentor

- 45,135

- 1,433

The net force on the 6 kg mass is: T1 - T2 (note the minus sign, since the tensions pull in opposite directions)So for the acceleration:

Fnetx of system=T=ma

T1+T2=6a

T1=6a-T2

So Newton II tells us:

T1 - T2 = 6a [equation 1]

That's good. Now apply Newton II:Fy= 117.6-T1 (for m=12kg)

117.6-T1 = 12a [equation 2]

Also good. Newton II:Fy= T2-39.2 (for m=4 kg)

T2-39.2 = 4a [equation 3]

Now combine those three equations and solve for the three unknowns. (Hint: Just add the equations together and see what happens.)

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