A box of 200kN is placed on a smooth inclined plane 28° with the horizontal. To prevent the box from sliding down the plane, a rope is tied to it and fastened to a peg on the top of the incline as shown. Determine the pull on the string assuming that the rope is held parallel to the plane. (Hint: the pull in the string must be equal to the component of the weight of the box parallel to the palne)
m*g = 200,000N.
Fp = mg*sin A = 200,000*sin28 = 98,894 N
= Force parallel with the incline = The
pull on the rope.
To determine the pull on the string, we need to find the component of the weight of the box parallel to the plane.
The weight of the box is given as 200 kN, which we can convert to Newtons by multiplying by 1000:
Weight of the box = 200 kN = 200,000 N
The weight of the box acts vertically downwards. To find the component of the weight parallel to the plane, we need to find the force acting perpendicular to the plane.
The force acting perpendicular to the plane can be found using trigonometry. The angle of the incline is given as 28° with the horizontal.
We can use the following trigonometric relationship:
sin(θ) = opposite/hypotenuse
In this case, the opposite side is the force acting perpendicular to the plane, and the hypotenuse is the weight of the box.
Using the given angle of 28°, we can substitute these values into the equation:
sin(28°) = Force perpendicular/Weight of the box
Rearranging the equation, we have:
Force perpendicular = sin(28°) * Weight of the box
Calculating this, we get:
Force perpendicular = sin(28°) * 200,000 N
Now, the pull on the string is equal to the force acting parallel to the plane, which is equal in magnitude to the force perpendicular to the plane.
So, the pull on the string is:
Pull on the string = Force perpendicular = sin(28°) * 200,000 N
Calculating this, we get the final answer for the pull on the string.