- #1

masterchiefo

- 212

- 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 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 * 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