An engine of mass 60 000kg is pulling a truck of mass 12 000kg at constant speed up a slope inclined at x to the horizontal, where sinx=1/200
Respective frictional resistances are 50N per 1000kg for the engine
a)the tractive force exerted by the engine
b)The tension in the coupling between the engine and the truck
Ans: a) 6888n b) 948N
2. A particle of mass 10kg rests in limiting equilibrium on a rough plane inclined at an angle of 30degrees with the horizontal. The plane is raised until its slope is 60defgrees. Find the magnitude of the additional force parallel to the plane required to support the body.
1 Answer | Add Yours
1. In the first problem the frictional coefficient of the truck is not provided.
2. A particle of mass 10 kg rests in limiting equilibrium on a rough plane inclined at an angle of 30 degrees with the horizontal.
The gravitational force of attraction on the particle is 10*g N acting vertically downwards. The force pulling the particle down is 10*g*sin 30. The resistive frictional force is 10*g*cos 30*C.
At equilibrium, 10*g*sin 30 = 10*g*cos 30*C
=> C = tan 30
When the plane is raised such that the angle made with the horizontal is 60 degrees, the downward force is 10*g*sin 60 and the resistive frictional force is 10*g*cos 60* tan 30
The additional force that needs to be applied is support the particle and prevent it from moving downwards is 10*g*sin 60 - 10*g*cos 60* tan 30
= 98*(sin 60 - cos 60*tan 30)
~~ 56.58 N
The force that has to be applied is 56.58*cos 60*i + 56.58*sin 60*j
We’ve answered 318,967 questions. We can answer yours, too.Ask a question