box with static friction?

Question

a 68 kg crate is dragged with a massless rope at an angle of 15 degrees above the horizontal. if the coefficient of static friction is .5....what is the min. force required to start the rate moving. also...if kinetic friction is .35, what is the magnitude of the initial acceleration of the crate? Please help me!

can anybody show me how to work this...i keep getting 320 N for the first part

supposidly the answer is suppose to be 300N and 1.3 m/s^2....now how is that?

Answer 1

when the crate is starting to move, the friction force would be equal to 0.5R...R is the ground reaction to the box.
calculate sigma F vertical

Fsin15 + R = 68g (g is the gravitational acceleration)

F = (68g/Sin15) - (R/Sin15) --- (1)

calculate Sigma F horizontal

FCos15 = 0.5R

R = (FCos15/0.5)

substitute with the value of R at (1) and solve it for F, thus you got it.

for the acceleration thing, it turns into Dynamics;

so Sigma F horizontal = ma (a is the requested acceleration)

Fcos15 - (0.35R) = 68a --- (2)

you have F and R (calculated them above)...substitute and solve the above equation for a

Edit :
Okay I'll solve it to you in numbers..
after substitutio;
F = (68g/sin15) - (2Fcot15)
therefore
F (1 + 2cot15) = (68g/sin15)
F = ((68g/sin15)/(1+2cot15))
use a calculator
you get F = ~ 304.199 N

R = 2Fcos15
R = ~ 587.667 N

now sub. at (2) ;

Fcos15 - 0.35R = 68a

a = ((Fcos15)/68) - ((0.35R)/68) = ~1.296 m/s^2

Quod Erat Demonstrandum !

Answer 2

Fw=mg
Fw=(68)(9.8)= 666.4N

Fw(x)=sin(15)666.4N=172.47
Fw(y)=cos(15)666.4N=643.69N

Edit: you know what, i think i did something wrong. Oh yeah! the normal force, it is not 643.69N. It's supposed to be FN= 666.4-172.47=493.93N
Friction=(mu)(FN)
Friction=.5(491.93)
Friction= 246.97N
246.97N is the mininum force.

Friction= mu(FN)
Friciton= .35(493.93)
Friction= 172.87

a= Fapply-Friciton(k) / m
a= 246.97N-172.87N / 68kg
a= 1.08m/s^2

Answer 3

1.99 ft. lbs.