A block of mass 4.20 kg is pushed up against a wall by a force P that makes an angle of θ = 50.0° angle with the horizontal as shown below. The coefficient of static friction between the block and the wall is 0.300.
|Pmax|
|
= N |
|Pmin|
|
= N |
(a)
The minimum force exerted on the block to avoid slide and the friction force will acts upwards.
Substitute the values,
P sin+Nmg Р яiл 0+ пP_ соs0 - тg Р- (sin @ + дсоsе)- тg = mg sin@ + дсose (4.20kg)(9.8m/s2) sin50° +(0.300)cos(50°) Р. =42.9N 43N
The maximum force exerted on the block to sliding upwards and the fr...
P.sn0- μΝ =m P.sinθ-μP.. cs0- mg P(nθ- μοs 0) = mg πag Р. sin θ- μαs θ (4.20ks) (9.3π/s") Р. sin50°-(0300)cos(50°) -71.8N 72N
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