
Starting from rest, a flat car is given a constant acceleration ${{a}_{o}}=2m/{{s}^{2}}$. A cable is connected to a crate A of mass $50kg$ as shown. Neglecting the friction between floor and car wheels and mass of pulley. The coefficient of friction between the crate and floor of the car is $\mu =0.3$. The tension in the cable is
Answer
557.1k+ views
Hint: In order to solve the question we have to consider the concept of equilibrium of force. We should also have to keep in mind that the frictional force is in the opposite direction of the motion. As given in the figure we can clearly see that the wire is going round the pulley and its acceleration will be twice that of the cart (pulley) as the wire does cross over the pulley.
Formula Used: We are going to use the following formula to solve the problem:-
$T-f=ma$.
Complete step-by-step solution
From the question above we have the following parameters with us:-
Acceleration of car, ${{a}_{o}}=2m/{{s}^{2}}$.
Acceleration of the crate $a=2\times 2=4m/{{s}^{2}}$ because the wire is going round the pulley and its acceleration will be twice as that of the cart because the wire does cross over the pulley.
Mass of crate A, $m=50kg$
We have to calculate the tension, $T$ in the cable.
We have the following figure with us:-
The free-body diagram of the block will be as follows:-
From the free body diagram and according to the question normal reaction (NR) and weight (mg) will balance each other.
Therefore, the net force on the block will be given as follows:-
$T-f=ma$………….. $(i)$where $f$ is the frictional force on the block.
But, $f=\mu N$………………… $(ii)$
Equation $(ii)$ is the formula of frictional force. The frictional force is the product of the coefficient of friction and normal reaction.
Putting $(ii)$in $(i)$ we get
$T-\mu N=ma$……………. $(iii)$
Putting the values of the parameters we get
$T-50\times 0.3\times 10=50\times 4$ (as $N=mg$)
$\Rightarrow N-150=200$
$\Rightarrow T=200+150$
$\Rightarrow T=350N$
Therefore the tension in the cable will be, $T=350N$
Note: We have to take care of the acceleration of the crate as it is twice that of the car because the wire does cross over the pulley. Drawing a correct free body diagram is a very important aspect of solving these types of problems. We can also take the value of $g=9.8m{{s}^{-2}}$ .
Formula Used: We are going to use the following formula to solve the problem:-
$T-f=ma$.
Complete step-by-step solution
From the question above we have the following parameters with us:-
Acceleration of car, ${{a}_{o}}=2m/{{s}^{2}}$.
Acceleration of the crate $a=2\times 2=4m/{{s}^{2}}$ because the wire is going round the pulley and its acceleration will be twice as that of the cart because the wire does cross over the pulley.
Mass of crate A, $m=50kg$
We have to calculate the tension, $T$ in the cable.
We have the following figure with us:-
The free-body diagram of the block will be as follows:-
From the free body diagram and according to the question normal reaction (NR) and weight (mg) will balance each other.
Therefore, the net force on the block will be given as follows:-
$T-f=ma$………….. $(i)$where $f$ is the frictional force on the block.
But, $f=\mu N$………………… $(ii)$
Equation $(ii)$ is the formula of frictional force. The frictional force is the product of the coefficient of friction and normal reaction.
Putting $(ii)$in $(i)$ we get
$T-\mu N=ma$……………. $(iii)$
Putting the values of the parameters we get
$T-50\times 0.3\times 10=50\times 4$ (as $N=mg$)
$\Rightarrow N-150=200$
$\Rightarrow T=200+150$
$\Rightarrow T=350N$
Therefore the tension in the cable will be, $T=350N$
Note: We have to take care of the acceleration of the crate as it is twice that of the car because the wire does cross over the pulley. Drawing a correct free body diagram is a very important aspect of solving these types of problems. We can also take the value of $g=9.8m{{s}^{-2}}$ .
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

