- #1
masterchiefo
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- 2
Homework Statement
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 .
a) Will the block slide? ( justify your answer )
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.
Homework 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.
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 secProblem 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 * axBlock 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