- #1
crimsonn
- 30
- 0
1. A 28.0 kg block is connected to an empty 1.35 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.450 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move. (a) Calculate the mass of sand added to the bucket. (b) Calculate the acceleration of the system
2. F=ma Ffr = mumg
3. I think I got part a.
For a)
mu(static)mg + mu (kinetic)mg =mg
(.45)(28)(9.8) + (.32)(28)(9.8)= m (9.8)
211.288 = m (9.8)
m= 21.58
subtracting 1.35 (mass of bucket)
mass of sand = 20.21
for b) You look at only the horizontal forces on the block, because vertical forces are going to cause any acceleration?
So, the sum of the forces on the box (1) is
m1a= Tension - Ffr
And on box 2
m2a = T - m2g
the acceleration is going to be the same because it is all attached, correct?
If my equations are correct, I'm just kind of confused about how to put them together.
help is really appreciated! thank you!
2. F=ma Ffr = mumg
3. I think I got part a.
For a)
mu(static)mg + mu (kinetic)mg =mg
(.45)(28)(9.8) + (.32)(28)(9.8)= m (9.8)
211.288 = m (9.8)
m= 21.58
subtracting 1.35 (mass of bucket)
mass of sand = 20.21
for b) You look at only the horizontal forces on the block, because vertical forces are going to cause any acceleration?
So, the sum of the forces on the box (1) is
m1a= Tension - Ffr
And on box 2
m2a = T - m2g
the acceleration is going to be the same because it is all attached, correct?
If my equations are correct, I'm just kind of confused about how to put them together.
help is really appreciated! thank you!