Can someone show me how to solve this physics equation for me?

1. First draw out the diagram and the direction of force you can draw on the block.




2. Then , you need to know one thing that the force pushing the block downwards is equal to the frictional force applied between the box and the ramp.

therefore, force = frictional force




3. Find the force first, applying Newton's First law:

F=ma
=(30cos20°)(1.2)
=33.83N

I used cos20° because the force applied on the block is resolved into two sides which can be found using sin? and cos?. This one can be explained through the diagram as well.

In other words,

F = Fsin? + Fcos?




4. Now that you have the force equals to 33.83N,

force = frictional force
therefore,

frictional force = 33.83N




5. Applying formula,

Frictional force, F = R?

where R is normal reaction and ? is the coefficient.




6. Find the normal reaction first,

R = mg
=(30)(9.81)
=294.3N




7. Substituting into the formula,

Frictional force, F = R?
33.83N = (294.3N)(?)
? = 0.1150

Hope its brief enough

kinetic friction = (coeff of frc.) * normal force = f
downward force on block is (mgsin?) and upward will be friction. (if we take the floor of the ramp as x-axis)
see by rotating the plane
net force = mgsin? - f = ma
sin20=0.9
normal force = mg
30*10*0.9 -(coeff. ) 30*10 = 30*1.2
10(0.9 - coeff) = 1.2
coeff = 0.9- (1.2/10)
=0.78

The normal reaction is R=30cos20 x 9.8N=276N; the friction force up the incline is ?R=276?N, the gravity force down the incline is 30 sin20 x 9.8N=101N.But the net force on the block =ma=30x1.2N=36N. So (101 - 276?)N=36N, giving ?=65/276=0.24.