â¢A block of mass 2.0 kg initially at rest on a horizontal floor moves under the action of an applied horizontal force
Equilibrium A single force acting on a body produces a change in both its translational and rotational motion
When several forces act on body simultaneously with the result that there is no change in either the translational or rotational motion, the body is said to
be in equilibrium
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The first condition of equilibrium It states that for a body to be in equilibrium, the vector sum of all the forces acting on that body must be zero or i.e F 0 Fx 0 Fy 0 Fz 0 • Static equilibrium • Dynamic equilibrium
2
θ
T T
θ F
`
w w Tx = - T sin θ Ty = T cos θ
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FORCES
Application of stable equilibrium
• A block of mass 2.0 kg initially at rest on a horizontal floor moves under the action of an applied horizontal force of 10 N. If the coefficient of kinetic friction between the block and the floor is 0.2, find the acceleration.
R fk
F
W
m = 2. 0 kg, W = mg = 20 N W = 20 N, μk = 0.2, F = 10 N, a = ?
Σ FY= 0 R–W=0 R = W = 20 N Σ FX = m a F – fk = m a F – μk R = m a a=
𝟏𝟎 −( 𝟎.𝟐 𝒙 𝟐𝟎 ) 𝟐
= 3 m s-2
• An object weighing 300N is required to move with a constant velocity down an inclined plane whose of inclination is 30°. Find the force which must act on the object, parallel to the plane. The coefficient of kinetic friction between two contact surfaces is 0.75.
An object is in dynamics equilibrium, 𝛴FY = 0 R – W cos 30 = 0 R = 300 cos 30 = 259.8N fk
R
F
𝛴FX = 0 fk – W sin 30 –F = 0 𝝁k R – 300 sin 30 = F F = 44.9 N
wcos
w
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• Suppose the velocity of the blood flowing in an artery was found to be 120 cms-1 at one point and 80 cms-1 at a point 50 cm further along the artery. Assuming constant deceleration, how long would it take the blood to flow from one point to the other? Also find the deceleration of the blood.
• An object weighing 200 N is in static equilibrium on a 30° inclined plane. Find fs and R.
fs
R
30
W sin 30
30 ̊ 200 N
W cos 30
• W = 200 N, fs = ?, R = ? An object is in static equilibrium, Σ Fy = 0 R – W cos 30 = 0 R = 200 cos 30 = 200 x 0.866= 173.2 N
Σ FX = 0 fs = W sin 30 fs = 200 sin 30 = 100 N
• An object weighing 100 N is required to move with a constant velocity up an inclined plane whose angle of inclination is 30°. Find the force, which must act on the object, parallel to the plane. The coefficient of kinetic friction between two contact surfaces is 0.15.
F
R W sin 30
fk 30 ̊
30
100 N W = 100 N, μk = 0.15
W cos 30
• An object is in dynamic equilibrium, Σ Fy = 0 R – W cos 30 = 0 R = 100 cos 30 = 100 x 0.866 = 86.6.N Σ FX = 0 F- fs - W sin 30 = 0 F =fs+200 sin 30 = μk R + 100 = 0.15 x 86.6 + 100= 113 N
• A 50 lb box is to be moved across a rough floor. The coefficient of static friction is 0.5 and that of kinetic friction is 0.3. How much force is required to start the box in motion and how much force is required to keep it moving with a constant speed?
R fs
F
W
W = 50 lb, μs = 0.5, F = ?
• The box is in static equilibrium, Σ Fy= 0 R–W=0 R = W = 50 lb Σ FX = 0 F – fs = 0 F = fs = μs R = 0.5 x 50 = 25 lb
μ = f/R
• The box is in dynamic equilibrium, Σ Fy= 0 R–W=0 R = W = 50 lb Σ FX = 0 F – fk = 0 F = fk = μk R = 0.3 x 50 = 15 lb
μ = f/R
