A block weighing 6 kg6kg is on a plane with an incline of (5pi)/65π6 and friction coefficient of 4/545. How much force, if any, is necessary to keep the block from sliding down?

1 Answer
Jan 3, 2016

0.8N up the slope.

Explanation:

Start by setting your problem up in a diagram like so:
My own imageMy own image
(gg is the acceleration due to gravity, value =9.81ms^-2=9.81ms2)
Then think about which formulae and physics concepts you will need to solve the problem.

We have the idea of equilibrium. We want the forces to balance so that the block doesn't move. So any resultant force should be zero.

We are mostly concerned with the forces parallel to the slope. Let's calculate the friction.

F=muRF=μR

Where FF is friction, muμ is the coefficient of friction, and RR is the normal reaction force.

So from my diagram we get the result:
F=2/5NF=25N

To find NN we resolve forces perpendicular to the slope:

N=6gcos((3pi)/8)N=6gcos(3π8)

So now we have

F=2/5*6gcos((3pi)/8)F=256gcos(3π8)

Now we will resolve parallel to the slope, remembering that we want the resultant force to equal zero, so we will need to introduce an additional force to balance it out.
I will call it TT, and define it as acting up the slop, in the same direction as FF.

F+T-6gsin((3pi)/8)=0F+T6gsin(3π8)=0

2/5*6gcos((3pi)/8)+T-6gsin((3pi)/8)=0256gcos(3π8)+T6gsin(3π8)=0

T=6gsin((3pi)/8)-2/5*6gcos((3pi)/8)T=6gsin(3π8)256gcos(3π8)

T=6g(sin((3pi)/8)-2/5cos((3pi)/8))T=6g(sin(3π8)25cos(3π8))

Numerical answer, T=0.770806....
T=0.8N (1 s.f.)