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2 answers:
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Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d= * tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d= * tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip
Answer:
(a) Acceleration of the bag will be a=16.214m/sec^2
(B) Weight of the bag will be 137.2 N
Explanation:
We have given mass of the bag m = 14 kg
Force with which bag is lifted = 227 N
(A) According to newtons law we force is equal to F = ma , here m is mass and a is acceleration
So
(b) Acceleration due to gravity
We know that weight is given by W = mg , here m is mass and g is acceleration due to gravity
So weight
So weight of the bag will be 137.2 N