Verified answer

Well, Here's the solution.

Let T1 be the tension on the string between the wall and m1.

Let T2 be the tension on the string between m1 and m2.

Let F1 be the frictional force between the ramp and m1.

Let F2 be the frictional force between the ramp and m2.

Let N1 and N2 be the normal reactions on m1 and m2 respectively.

Let C be the coefficient of friction.

Now,

F1 = CxN1 and F2 = CxN2

N1 = m1 g sin35 and N2 = m2 g sin35

At equilibrium,

m2 g cos35 = F2 T2

m1 g cos35 T2 = F1 T1

There! You can find T1 and T2 using the above two equations.

Enjoy!

because of the fact the coef. of static friction is given, it fairly is nicely worth checking to work out if the equipment will pass in any respect .. Friction between m1 and table .. Ff = ?(s).R .. .. (R = perp. stress, m1g) Ff = 0.40 4 x (41kg x 9.80) .. .. .. Ff = 176.80 N whilst the equipment is at relax the stress interior the string, T = m2.g (weight of m2) T = 15kg x 9.80 .. .. T = 147.0 N the stress isn't adequate to go m1 from relax .. ?T continues to be at 147.0 N