Ice block on inclined plane- easy problem

[cep]

Ice block on inclined plane-- easy problem!

I'm pretty sure I did this correctly, but my answers are different from those given in the back of the book (which our TA warned us were frequently incorrect), so I thought I'd ask you guys if I made any obvious mistakes.

Homework Statement

A large block of ice of mass M=80kg is held stationary on a frictionless ramp. The ramp is at an angle of theta=36.9º above the horizontal.

a) If the ice block is held in place by a tangential force along the surface of the ramp (at angle theta to the horizontal), find the magnitude of the force.

b) If, instead, the ice block is held in place by a horizontal force, directed horizontally towards the center of the ice block, find the magnitude of this force.

Homework Equations

Newton's laws

The Attempt at a Solution

a) I rotated my axes by 36.9º so that the x-axis is parallel to the slope of the incline. Only the x-components matter in this problem, so I'm going to ignore the y-axis on here. For no acceleration to take place, F-mgx=0; F=mgx. In this case, mgx=mgsin(theta). F=80*9.8*sin(36.9)=470 N. The book's answer is 500 N, which is the same answer with 1 sigfig.

b) Keeping the same rotated axes, a horizontally applied force is not parallel to the axes. So, the relevant component of the force is Fx=Fcos(theta). Again, Fx-mgx=0 for no acceleration to take place. Fcos(theta)-mgsin(theta)=0

Fcos(theta)=mgsin(theta)
F=mgtan(theta)=80*9.8*tan(36.9)=590 N. The book's answer is 400 N, which doesn't make much logical sense to me, but maybe you all see where they're coming from. Thank you!

-CEP

 

[Doc Al]

Fcos(theta)=mgsin(theta)
F=mgtan(theta)=80*9.8*tan(36.9)=590 N. The book's answer is 400 N, which doesn't make much logical sense to me, but maybe you all see where they're coming from.

The book's wrong, but you are correct. (I guess this is one of those times the TA warned you about.)

 

[cep]

Thanks, Doc!