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M1 question - too difficult - please help

1. A small parcel p, of mass 1.5 kg, is placed on a rough plane inclined at an angle of 27 deg to the horizontal. The coefficent of friction between the parcel and the plane is 0.3. A force S, of variable magnitude, is applied to the parcel. The line of action of S is parallel to the line of greatest slope of the inclined plane. Determine in newtons to 1 decimal place, the magnitude of S when the parcel P is in limiting equilibrium and on the point of moving:

a) down the plane b) up the plane

2. A small canister is attached to a helium filled balloon and released from rest at ground level. After 4 secs, it is moving vertically upwards at 6m/s.

a) Find the height of the balloon and canister after 4 secs, stating clearly any assumptions you make.

b) When the balloon reaches a height of 27m, it bursts. Find the maximum height reached by canister
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1. parcel p, mass 1.5 kg, placed on a rough plane inclined at 27 deg to the horiz. coefficent of fric is 0.3. Force S, variable magnitude, is applied to the parcel. The line of action of S is parallel to the line of greatest slope of the inclined plane. Calc. magnitude of S when p is in limiting equilibrium and on the point of moving:

a) down the plane b) up the plane

R = 1.5gcos27 = 13.1 N
Friction = uR = 0.3(13.1) = 3.93 N (3.S.F)

Down The Plane is defined as the +ve direction.
Up The Plane is defined as the -ve direction.

a.) Resolving Parallel to the plane:

1.5gsin27 - 3.93 + S = 0
---> S = 3.93 - (1.5)(9.8)(sin27)
---> Magnitude S = |-2.744|
---> Magnitude S = 2.74 N (3.S.F)

b.) Resolving Parallel to the plane:

-1.5gsin27 - 3.93 + S = 0
---> S = 3.93 + 1.5gsin27
---> Magnitude S = 10.6 N (3.S.F)

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2. A small canister is attached to a helium filled balloon and released from rest at ground level. After 4 secs, it is moving vertically upwards at 6m/s.

a) Find the height of the balloon and canister after 4 secs, stating clearly any assumptions you make.

b) When the balloon reaches a height of 27m, it bursts. Find the maximum height reached by canister

a.) t = 4, v = 6, u = 0.
s = [(u + v)/2] * t
---> s = [6/2] * 4
---> Height (At t = 4) = 12 m

Assumptions: The canister is modelled as a point mass and hence air resistance against the canister is deemed negligible.
Negligible air resistance acts on the balloon during its motion.


b.) At point where balloon bursts: a = -9.8, u = 6, v = 0.
v = u + at
---> 0 = 6 - 9.8t
---> t = 0.612s

s = ut + 1/2at^2
---> s = 6(0.612) + 1/2(-9.8)(0.612)^2
---> s = 1.84 m

Maximum Height Reached = 27 + 1.84
= 28.8 m (3.S.F)

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