- #1

mrjeffy321

Science Advisor

- 876

- 1

Here is the question:

Block A:

mass = 4 kg - given

the normal force = 34 N - normal force = m*g*cos(angle)

the force (parralell to the ramp surface) fulling it down

=19.62 N - force paralell = normal force * sin (angle)

no frictional force

Block B:

mass = 2 kg - given

normal force = 19.62 N - formal force = m*g

coefficent = kennetic friction = .5 - given

force of kinetic friction = 9.81 N - force of friction = normal force * coefficient of friction

so now there is a 19.62 N force pulling the blocks down, and a 9.81 N force trying the pull block B back, so as I see it, the tension in the cord is equal to 16.62-9.81 = 9.81 N. and then since there is a 9.81 N force acting on both blocks, the acceleration = net force / total mass = 1.64 m/s^2.

I got the acceleration correct, says the book, but not the tension force.

If I look in the backl of the book, it gives me the answer, which is not what I calculated.

A. Tension in the cord = 13 N

B. Acceleration = 1.6 m/s^2 ___I got this one right

I would really like to understand how to do this problem correctly.

I have attached the picture of the situation and below is how I tried to solve it.Block A in [figure that I attached] has a mass of 4.0 kg, and block B has a mass of 2 kg. The co-efficient of kinetic friction between block B and the horizontal plane is .50. The incline plane is frictionless and at an angle of 30 degrees. The pulley serves only to change the direction of the cord connecting the blocks. The cord has negligible mass.

Quesitons:

A. Find the tension in the cord

B. Find the magnitude of acceleration of the blocks.

Block A:

mass = 4 kg - given

the normal force = 34 N - normal force = m*g*cos(angle)

the force (parralell to the ramp surface) fulling it down

=19.62 N - force paralell = normal force * sin (angle)

no frictional force

Block B:

mass = 2 kg - given

normal force = 19.62 N - formal force = m*g

coefficent = kennetic friction = .5 - given

force of kinetic friction = 9.81 N - force of friction = normal force * coefficient of friction

so now there is a 19.62 N force pulling the blocks down, and a 9.81 N force trying the pull block B back, so as I see it, the tension in the cord is equal to 16.62-9.81 = 9.81 N. and then since there is a 9.81 N force acting on both blocks, the acceleration = net force / total mass = 1.64 m/s^2.

I got the acceleration correct, says the book, but not the tension force.

If I look in the backl of the book, it gives me the answer, which is not what I calculated.

A. Tension in the cord = 13 N

B. Acceleration = 1.6 m/s^2 ___I got this one right

I would really like to understand how to do this problem correctly.