How Much Force Is Needed to Move a 59 kg Crate?

[anthonyr]

I have a homework due to this saturday, and i solved all the problems except for the following 4, if you please solve them, and post them with the solution, ill be greatful:D thanks in advance:) :

1)A 59 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 12° above the horizontal. (a) If the coefficient of static friction is 0.56, what minimum force magnitude is required from the rope to start the crate moving? (b) If μk = 0.33, what is the magnitude of the initial acceleration (m/s^2) of the crate?

2)Block A in Fig. 6-57 has mass mA = 4.00 kg, and block B has mass mB = 2.10 kg. The coefficient of kinetic friction between block B and the horizontal plane is μk = 0.510. The inclined plane is frictionless and at angle θ = 32.0°. The pulley serves only to change the direction of the cord connecting the blocks. The cord has negligible mass. Find (a) the tension in the cord and (b) the magnitude of the acceleration of the blocks. (image is attached)

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/pict_6_57.gif

3)The force on a particle is directed along an x-axis and given by F = 0.96(x/2.6 - 1.6) where x is in meters and F is in Newtons. Find the work done by the force in moving the particle from x = 0 to x = 7.7 m4)A horse pulls a cart with a force of 37 lb at an angle of 29° above the horizontal and moves along at a speed of 7.2 mi/h. (a) How much work (in ft lb) does the force do in 23 min? (b) What is the average power (in horsepower, hp) of the force?

Thanks in advance:D

Homework Statement

Homework Equations

The Attempt at a Solution

answers:

1) coefficient of static friction=0.56, then friction f= 0.56*m*g = 0.56*59*9.8=323.792N

at that stage, the system is in equilibrium, then summation of forces=0, then f=Tx= 323.792N

=>T, which is equal to Tx/cox(teta) where teta=12degrees, T=331.025N.

therefore, minimum force magnitude required to move the crate = 331.025N.

b)f2 = 190.8N = 0.33 * 59*9.8
Summation forces = Ma, where ay=0 and ax = (T-f2)/M = 2.254 m/s^22)P = m*g = 4*9.8 =39.2N
Px = Psin32 = 20.77 N = T
=> Tx = Tcos 32 = 17.61N

On B, f = 10.5N = Tb

Tequ = 17.66-10.5=7.16N

b) Summation forces = Ma, => 7.16/6.1 = a => a=1.17 m/s^23)W = F.(variation of x) = |F|.|x|.cos(teta) where teta=0 because F and x have the same orientation.
=> W = 0.96(x/2.6-1.6) * (x-x0) => W = 10.06 joules. where x-x0=7.7m

4) I didnt even know from where to start because I am not used to US units... I tried converting everything to SI, but still it didnt work!
2)

 

[willem2]

For question 1, since the rope pulls upwards also, this will reduce the normal force and so the friction is also reduced

 

[anthonyr]

so what will the answer be? and what about the rest? :(

 

[marcusl]

Anthonyr, you are required by the rules of this forum to attempt to solve the problems yourself, and to post your work. Then we can help you where you got stuck.

 

[anthonyr]

I did solve the problems, and posted them under answer, but the answers aren't correct... there is only the 4th question that i didnt really know... please someone answer my questions... I am sure i tried to do them as i posted my attempt to answer them... thanks:)