Question

In Figure 2, a block of weight W=100 N is sliding down a 2-stage ramp defined by the angles a = 30° and B = 45°. The distances OA and OB are 2 m and 4 m respectively. The friction coefficient for the entire ramp is u=0.3. B Figure 2: System for Question 5. (a) Derive an expression for the component of the block's weight acting parallel to the 2-stage ramp (Hint: since the ramp has 2 stages defined by separate angles, your expression will be a piecewise function) (b) Use an appropriate force balance to derive an expression for the reaction force acting on the block as it slides down the ramp (Hint: again, the result will be piecewise). (c) Use your answer in (b) to write down an expression for the frictional force acting on the block as it slides down the ramp.

Answer 1

111 Page ① oli Given values in question are : weight of block (W)=loon Q = 30° B = 45° OA = 2m OB = 4m friction coefficient for entire ramp = ju=0.3 a) Block's weight acting parallel to first ramp = mg sind = wsind (since wang) mgogoemas = loux sin 30° = 100x] -50N mgsing " mg ao Hence, Block's weight acting parallel to first ramp is son. Block's weight acting parallel to second ramp = mgsinß = Wsinß 100x Sin 45 mgressen I loox = 70.71N parallel to second ramp is 70.71N Hence, Block's weight acting . (b) Reaction for 0 Reaction force acting on block as it slides down the first ramp = N1 = Mg Cosx Ni =ing Cosa (By force balance) Ny = Wash Ni - 1oox Cos 30° Ni=86.600 Hence, Reaction force acting on block as it slides down the first ramp is 86-60N. mg cost Asmasina ng 20 Reaction force acting on block as it slides down the second ramp = N2 = mg aos N2 = mg cosp (By force balance) mg cosplay W2=w Cosp N2 = 100x Gs 45° N2=70.71N Hence, Reaction force acting on block as it slides down the second ramp is 70-71N. -

(C) Friction force Page ② friction force acting on block as slides down the first ramp = fl=UN, fp = MN1 fj = Mx 8G60N (No calculated in part (6)) fx = 0.3486.60 fx = 25.980 macoso ng 20 Hence, friction force acting on block as it slides down the first ramp is 25-98N Forction force acting on bloc as it slides down the second ramp = fz = lN2 fa part (b)) mg cose mgsinß f2= MN2 f2 = MX 70.71 (Nz calculated in f2 = 0.3870.71 fz z 21-21N • Diction force acting on block as it slides down the second ramp is 21:21N.
This block ramp problem which consists of two ramps, therefore, we need to find out block's weight parallel to ramp, reaction force and friction force for both the ramps individually.