A crate of mass 106 kg is pulled by a rope on a frictionless surface. The rope makes an angle of 40 0 above the horizontal. If the tension on the string is 1183 N, what is the acceleration of the crate?

Answer 1

1183*cos40=106*acceleration
your teacher is too easy, should have added friction.

Answer 2

Draw a FBD

then,

find component of forces in the x and y direction:

F_x=1183*cos(40) = 906.231 N

F_y=1183*sin(40)-1039.86 = -279.442 N

Now find the resultant force:

F_r = sqrt(F_x^2 + F_y^2)
F_r = sqrt(906.231^2 + (-279.442)^2)
F_r = 948.337 N

a = F_r / m

a = 948.337 / 106 = 8.94657 m/s^2

Answer 3

Don: I don't think the mass leaves nor goes below the surface because of weight and the surface is hard, so the only acceleration can be horizontal, no component downward. If the tension were greater, it could have went upward.

Answer 4

Whoa you're right, I read proble wrong did this as if the mass was on an inclined, but only the force is