
Akhil has to lift a load of 800N onto a platform of 2m height. Instead, he pushes the load up a ramp 4m long. Find the force required to roll the load up the ramp.
A. 600N
B. 800N
C. 300N
D. 400N
Answer
487.5k+ views
Hint: Draw the free body diagram of load on the ramp and draw the vertical and horizontal components of load mg. Balance all the loads and find the value of force required to roll the load up the ramp.
Formula used:
$\text{F=mgsin }\!\!\theta\!\!\text{ }$
Complete step by step solution:
First, we will draw the free body diagram of load, as shown below
Fig. Free body diagram
Now, we will calculate the angle $\text{ }\!\!\theta\!\!\text{ }$ as $\text{ }\!\!\theta\!\!\text{ =si}{{\text{n}}^{\text{-1}}}\text{(}\dfrac{\text{2}}{\text{4}}\text{)}$
So, we get $\text{ }\!\!\theta\!\!\text{ =3}{{\text{0}}^{\text{o}}}$
Here, mg is the load
$\Rightarrow \text{mg=800N}$
Also, $\text{mgsin }\!\!\theta\!\!\text{ }$ and $\text{mgcos }\!\!\theta\!\!\text{ }$ are the horizontal and vertical components of load respectively.
Therefore, we need to overcome the force $\text{mgsin }\!\!\theta\!\!\text{ }$ to roll the load up the ramp. The vertical component of load $\text{mgcos }\!\!\theta\!\!\text{ }$ will be balanced by the normal reaction of the body on the ramp.
Thus, we will require a force $\text{F=mgsin }\!\!\theta\!\!\text{ }$ to roll the load up the ramp.
Therefore, $\text{F=800sin(3}{{\text{0}}^{\text{o}}}\text{)}$
$\Rightarrow \text{F=800 }\!\!\times\!\!\text{ }\dfrac{\text{2}}{\text{4}}$
$\therefore \text{F=400N}$
Therefore, the force required to roll the load up the ramp will be 400N.
So, the correct answer is “Option D”.
Additional Information: Free body diagrams are used to picture the forces and moments acting on a body. It is used to calculate the resulting reactions in many types of mechanics problems. The total external force acting on the object is obtained by using FBD in order to apply Newton's Second Law of motion on the object. A series of free bodies diagrams may be necessary to solve complex problems of mechanics.
Note: One must use a component resolving method and draw a free body diagram of the load whenever necessary. In some questions only mass of the body is given. Then one must multiply it with the acceleration due to gravity g.
Formula used:
$\text{F=mgsin }\!\!\theta\!\!\text{ }$
Complete step by step solution:
First, we will draw the free body diagram of load, as shown below

Fig. Free body diagram
Now, we will calculate the angle $\text{ }\!\!\theta\!\!\text{ }$ as $\text{ }\!\!\theta\!\!\text{ =si}{{\text{n}}^{\text{-1}}}\text{(}\dfrac{\text{2}}{\text{4}}\text{)}$
So, we get $\text{ }\!\!\theta\!\!\text{ =3}{{\text{0}}^{\text{o}}}$
Here, mg is the load
$\Rightarrow \text{mg=800N}$
Also, $\text{mgsin }\!\!\theta\!\!\text{ }$ and $\text{mgcos }\!\!\theta\!\!\text{ }$ are the horizontal and vertical components of load respectively.
Therefore, we need to overcome the force $\text{mgsin }\!\!\theta\!\!\text{ }$ to roll the load up the ramp. The vertical component of load $\text{mgcos }\!\!\theta\!\!\text{ }$ will be balanced by the normal reaction of the body on the ramp.
Thus, we will require a force $\text{F=mgsin }\!\!\theta\!\!\text{ }$ to roll the load up the ramp.
Therefore, $\text{F=800sin(3}{{\text{0}}^{\text{o}}}\text{)}$
$\Rightarrow \text{F=800 }\!\!\times\!\!\text{ }\dfrac{\text{2}}{\text{4}}$
$\therefore \text{F=400N}$
Therefore, the force required to roll the load up the ramp will be 400N.
So, the correct answer is “Option D”.
Additional Information: Free body diagrams are used to picture the forces and moments acting on a body. It is used to calculate the resulting reactions in many types of mechanics problems. The total external force acting on the object is obtained by using FBD in order to apply Newton's Second Law of motion on the object. A series of free bodies diagrams may be necessary to solve complex problems of mechanics.
Note: One must use a component resolving method and draw a free body diagram of the load whenever necessary. In some questions only mass of the body is given. Then one must multiply it with the acceleration due to gravity g.
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