# Kinetics Problems -- Blocks on inclined plane

1. Apr 4, 2015

### masterchiefo

1. The problem statement, all variables and given/known data
Problem 1:
A block of mass m = 15 kg from rest on an incline of 20 °. It is assumed that
static and dynamic friction coefficients are respectively μs =0.4 and μk = 0.2. a force P
is then applied (P = 30 N) on the block as shown in Figure 1 .

if yes
b ) Draw the diagram of forces and acceleration;
c) Determine the acceleration of the block ( size and orientation) ;
d) if the block reaches a speed of 10 m / s , how far he has come ;
e) then determine the time taken by the block to reach the speed of 10 m / s.

Problem 2:
Two blocks A and B, of masses mA = 10 kg and mB = 7kg , connected by a non-elastic cable slide over
a map as shown in Figure 2. If we apply on the block A a force P = 100 N ( see Figure 2)
and that the kinetic coefficient of friction between the blocks and the map is μk = 0.12 ,
a) draw the diagram of the forces;
determine :
b) the acceleration of two blocks;
c) the direction of movement of the two blocks;
d) the tension in the cable connecting the two blocks.

2. Relevant equations
I would love if you guys could verify if all my answer are correct.
thank you very much.

I am a little confused when to use μk and μs in both problems.

3. The attempt at a solution
P = 30N
m = 15kg
W = m*g = 147.15 N
us = 0.40
uk = 0.20

Problem 1)
My drawing of the problem with the block and the inclined plane:
http://i.imgur.com/PFw3GoE.png

a)
∑Fx = 0 = -cos(60)*P + cos(20)*Ff - sin(20)*N
0 = -cos(60)*30+ cos(20)*Ff - sin(20)*N
∑Fy = 0 = -W + sin(20)*Ff + cos(20)*N - sin(60)*P
0 = -147.15N + sin(20)*Ff + cos(20)*N - sin(60)*30

0 = -cos(60)*30+ cos(20)*Ff - sin(20)*N
0 = -147.15N + sin(20)*Ff + cos(20)*N - sin(60)*30
I did a solve of those 2 equations on my TI-Nspire calculator to find N and Ff.
Ff = 73.2096N
N = 157.559N

Fs = 0.4 * 157.559N = 63.0236N
if Fs < 73.309N then there is a movement and my block is sliding.
63.0236N < 73.2096N

b ) Draw the diagram of forces;
http://i.imgur.com/PFw3GoE.png

C)
Ffk = uk * N = 0.2 * 157.559 = 31.5118
∑Fx = m*ax
-cos(60)*P + cos(20)*Ffk - sin(20)*N = 15kg * ax
-cos(60)*30 + cos(20)*31.5118 - sin(20)*157.559 = 15kg * ax
ax = -2.78m/s^2

d)
Xf-Xi=X
vf^2 = vi^2 +2*a(Xf-Xi)
(-10)^2 = 0^2 + 2*(-2.78)*(X)
-17,98m

e) vf = vi + a(tf-ti)
-10 = 0 + (-2.78)*(tf-0)
3.59 sec

Problem 2)

a) draw the diagram of the forces;
http://i.imgur.com/37SvuCN.png

b) c) and d)
ay = 0 for both blocks

Block A:
∑Fy = m*ay
N-W+P*sin(25)= 10Kg *0 => N-(9,81*10Kg)+100N*sin(25)= 0
N = 55.7382N
Fk = uk * N => Fk = 0.12 * 55.7382N
Fk = 6.68858N

∑Fx = m*ax
-P*cos(25) + T + Fk = 10kg * ax
-100N*cos(25) + T +6.68858N = 10kg * ax

Block B:
∑Fy = m*ay
N-W*cos(20)= 7Kg *0
N-(7*9.81)*cos(20)=0
N = 64.5287N
Fk = uk * N => Fk = 0.12 * 64.5287N
Fk = 7.74344N

∑Fx = m*ax
-T+W*sin(20)+Fk = 7Kg *ax
-T+(7*9.91)*sin(20)+7.74344N = 7Kg * ax

Solve with TI-Nspire calculator to find ax and T:
-100N*cos(25) + T +6.68858N = 10kg * ax ---Block A
-T+(7*9.91)*sin(20)+7.74344N = 7Kg * ax ---Block B
T = 52.935N <====== answer for question d)
ax = -3.10072 <====== answer for question b)

c) since ax is negative we can assume that the direction of movement of the two blocks is to the left.

Thank you very much

2. Apr 5, 2015

### SammyS

Staff Emeritus
Only comments on Prob. #1 here.

Part (a) & (b) appear to be correct. The friction force needed to keep the block from sliding id more than the maximum value that static friction can provide.

From this point on, you need to use kinetic friction.

After this point you only find and then use the x-component of acceleration. With the coordinate system you have chosen, both ax and ay are non-zero.

3. Apr 5, 2015

### masterchiefo

Yes you are completely right man.
I have corrected that mistake on my paper work.