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Blueberrymuffim

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## Homework Statement

A force F is applied horizontally to a block A of mass mA =

0.600 kg so that it slides on the floor with acceleration g/3.

Block B of mass mB = 0.200 kg rests on top of A, but is kept

in place by a cord which is tied to the wall at 45.0°. The

coefficient of kinetic friction between the two blocks is μ1 =

0.500 and between block A and the floor is μ2 = 0.250.

(a) Draw a free-body diagram for each block.

(b) Find the tension in the cord.

(c) Find the normal force that block A exerts on block B.

(d) Find the normal force that the floor exerts on block A.

(e) Find the force F.

## Homework Equations

F=m.a fk = Mk*n

## The Attempt at a Solution

I tried to find the Tension by the components of the vector. Initially, I found that since the angle of the force is 45 degree, then I have an isosceles triangle for the components which each force will be the same in the component by in different direction which changes the sign (just assuming).

Then, I separate the forces of the Tension in components Fy and Fx acting in the block B.

Fy= S (force of the component y of tension) + n (normal force) - W (mass of B * gravity)

I assumed that Fy would be equal to zero because it is not moving in the vertical direction which means the sum of all the components of this net force will be equal to zero.

So, I decided to isolate S + n= 1.96.

So, S+n = weight.

n=1.96 N - S

Fx= fb (friction force) + S(force of the component x of tension).

Fx= Mk*n+S

Fx=0.5n - S

Fx = 0.5(1.96 - S) -S

Fx=0,98 - 0.5S - S

Fx = m.a by the given information we have Fx= 0.653N

S = -3.266 N