Question

hey guys !!!
Important question of 100 marks ..
solve this its urgent..

please it's urgent Guys need breif solution.

Answer 1

Heyy mate here is your answer...

Consider the following figure.


Consider the angle made by the ladder with respect ground be angle θ.
Hence
cosθ=1215cosθ=35
Hence
sinθ=45
From the figure the weight of the object will be concentrated at the point which is at distance of 5 m from the ground and the man is at a distance of 7.5 m from the ground.
As the ladder is resting on the ground, the net force acting on the ladder should be equal to zero.
Taking moment of all the forces about the point where the ladder is resting we get
S(15sinθ)=mg(5cosθ)+Mg(7.5cosθ)S(15)(45)=500(5)35+800(7.5)35         [take g= 10]12S=1500+360012S=5100S=425 N
As the net vertical force is also zero in order to be in equilibrium,
R=mg+MgR=500+800R=1300 N

Hope it will help you...

Answer 2

Note : Attachment is of the above user .

Consider the point made by the stepping stool with deference ground be angle θ.

Subsequently

cosθ=1215cosθ=35

Subsequently

sinθ=45

From the figure the heaviness of the article will be amassed at the point which is at separation of 5 m starting from the earliest stage the man is at a separation of 7.5 m starting from the earliest stage.

As the step is laying on the ground, the net power following up on the stepping stool ought to be equivalent to zero.

Taking snapshot of the considerable number of powers about the point where the stepping stool is resting we get

S(15sinθ)=mg(5cosθ)+Mg(7.5cosθ)S(15)(45)=500(5)35+800(7.5)35         [take g= 10]12S=1500+360012S=5100S=425 N

As the net vertical power is likewise zero so as to be in balance,

R=mg+MgR=500+800R=1300 N