Acceleration of a Block on an Inclined Plane

[riseofphoenix]

7. A block of mass m = 5.5 kg is pulled up a θ = 22° incline as in the figure with a force of magnitude F = 38 N.

(a) Find the acceleration of the block if the incline is frictionless

So I got this one right, and here is what I did...

From the free body diagram I drew:

F_net = (38 N) + (-mg sin 22)
ma = 38 - mg sin 22
a = (38 - mg sin 22)/m
a = [ 38 - (5.5)(9.81)(sin 22) ] / (5.5)
a = (38 - 20.211) / 5.5
a = 3.24 m/s²

(b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.12.

So...I know that

f_k = μ_kn
f_k = 0.12(F_appliedsin θ - mg)

QUESTION 1: So this is where I'm a little confused...Normal force should be UPWARDS, right? then would that mean that n = F_appliedsin θ + (- mg) after rotating my free body diagram so that I can see the axis better? Because I almost got the answer...

f_k = 0.12[38 sin 22 - (5.5)(9.81)]
f_k = -4.767 (the negative number makes sense since kinetic friction is OPPOSING the motion of the block which is moving up the incline.

So then I add it to the equation for acceleration at the top

ma = (38 N) + (-mg sin 22) + (-4.767)
a = [ 38 - (5.5)(9.81)(sin 22) - 4.767 ] / (5.5)
a = 13.022 / 5.5
a = 2.36 m/s²

This answer is wrong...Webassign says the answer is actually a = 2.15 m/s²

QUESTION 2: Where did I go wrong? Was it the normal force equation part?

 

[Doc Al]

So...I know that

f_k = μ_kn

Right.

f_k = 0.12(F_appliedsin θ - mg)

Not right.

QUESTION 1: So this is where I'm a little confused...Normal force should be UPWARDS, right? then would that mean that n = F_appliedsin θ + (- mg) after rotating my free body diagram so that I can see the axis better? Because I almost got the answer...

The normal force is 'normal'--perpendicular--to the surface. Figure out the normal force by analyzing force components perpendicular to the incline. They must add to zero.

 

[riseofphoenix]

Right.

Not right.

The normal force is 'normal'--perpendicular--to the surface. Figure out the normal force by analyzing force components perpendicular to the incline. They must add to zero.

No wait...
it would be

-mg + (-mg sin 22)?

 

[riseofphoenix]

Right.

Not right.

The normal force is 'normal'--perpendicular--to the surface. Figure out the normal force by analyzing force components perpendicular to the incline. They must add to zero.

It's just F cos θ, which is 38 cos 22...

But I still don't get the right answer when I multiply 0.12 by 38 cos 22, and then add 4.22 to this equation...

ma = 38 - 20.11

a = (38 - 20.11 + 4.22) / 5.5 doesn't give me 2.15 m/s²

Something else is wrong

 

[Doc Al]

It's just F cos θ, which is 38 cos 22...

The applied force F is parallel to the incline. It has no normal component.

The weight of the block, being vertical, will have a normal component.

 

[riseofphoenix]

The applied force F is parallel to the incline. It has no normal component.

The weight of the block, being vertical, will have a normal component.

So it's -mg cos 22?
Which is basically -(5.5)(9.81)(cos 22)...

f_k = 0.12[-(5.5)(9.81)(cos 22)]
f_k = -6.003

So,

a = [ 38 - 20.211 + (-6.003) ] / 5.5
a = 2.159638636

:)

Thanks!

 

[Doc Al]

So it's 5.5 cos 22?

Almost. What's the weight of the block?

 

[riseofphoenix]

Almost. What's the weight of the block?

It's just 5.5 kg isn't it?

 

[Doc Al]

It's just 5.5 kg isn't it?

No, that's the mass.

 

[riseofphoenix]

No, that's the mass.

oh...Weight = mg
Weight = -(5.5)(9.81) = -53.955 N

 

[Doc Al]

oh...Weight = mg
Weight = -(5.5)(9.81) = -53.955 N

Right. The weight = mg and acts downward.

 

[riseofphoenix]

Right. The weight = mg and acts downward.

Ohh ok!
Thanks a lot!