Acceleration of block down an incline

[aliaze1]

Homework Statement

Find the acceleration of block A when moving downward

Theta=40 degrees
A=103N
B=31N
Static coefficient = 0.56
Kinetic coefficient = 0.25

Homework Equations

F=ma
Normal Force * Kinetic coefficient = Friction Force

The Attempt at a Solution

From 103/9.8 I get a mass of ~10.5

Breaking 103 into components, 103sin40 and 103cos40, the first being perpendicular and the second being along the plane

103sin40*0.25=Friction Force

Tension is 31, so the sum of forces is:

(31+103sin40*0.25)-103cos40 = Fnet

Fnet/10.5 = Acceleration of A

is this correct?

Thanks!

 

[jumpman11372]

you have mixed up the normal force and the force going down the ramp

mgcos40 would be the FN ; use this to find the frictional force
mgsin40 would be the force with which it accelerates down the ramp

try writing some net force equations ; those would help :]

 

[thomate1]

Yes, your approach is correct. The equation F=ma can be used to find the acceleration of the block, where F is the net force acting on the block, m is the mass of the block, and a is the acceleration. In this case, the net force is equal to the sum of all the forces acting on the block, including the weight (mg), the normal force (N), and the friction force (Ff). Using the given values, we can write the following equation:

Fnet = mg - Ff - N

Since the block is moving downward, the normal force is equal to the weight of the block, so we can rewrite the equation as:

Fnet = mg - Ff - mg

Simplifying this equation, we get:

Fnet = mg(1 - μk)

Where μk is the kinetic coefficient of friction. Substituting the given values, we get:

Fnet = (10.5 kg)(9.8 m/s^2)(1 - 0.25) = 76.125 N

Finally, using F=ma, we can find the acceleration:

a = Fnet/m = 76.125 N / 10.5 kg = 7.25 m/s^2

Therefore, the acceleration of the block A when moving downward is 7.25 m/s^2.