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Question: In fig, if \({f_1}\), \({f_2}\) and \(T\) are the frictional forces on \(2{{kg}}\) block, \(3{{kg}}\...

In fig, if f1{f_1}, f2{f_2} and TT are the frictional forces on 2kg2{{kg}} block, 3kg3{{kg}} block and tension in the string, respectively, then find their values. Initially before applying the forces, tension in string was zero.

Explanation

Solution

To solve the problem related to the friction forces, first we have to draw the free body diagram of the given diagram then calculate the all necessary forces required to get the final answer.

Complete step by step answer: Friction is the force which resists the motion of any particle or it is force acting between two surfaces which are in contact with each other and are in motion. The friction force always acts in the opposite direction of the motion.
The formula to calculate the friction force is given as follows.
f=μNf = \mu N
Here, μ\mu is friction coefficients, ff is friction force, and NN is normal force acting on the object.
Draw the free body diagram of the given diagram.

Consider the 3kg3{{kg}} block.
Find the normal force NN as follows.
N=mgN = mg
By substituting 3kg3{{kg}} for mm, and 10m/s210{{m/}}{{{s}}^{{2}}} for gg in the equation N=mgN = mg, we get,
N=(3kg)(10m/s2) =30N N = \left( {3{{kg}}} \right)\left( {10{{m/}}{{{s}}^{{2}}}} \right)\\\ = 30{{N}}
Find the frictional force in 3kg3{{kg}} block.
f2=μN{f_2} = \mu N
By substituting 30N30{{N}} for NN, and 0.20.2 for μ\mu , we get,
f2=(0.2)(30N) =6N {f_2} = \left( {0.2} \right)\left( {30{{N}}} \right)\\\ = 6{{N}}
Balance the force on 3kg3{{kg}} block to find the tension (T)\left( T \right).
f2+T=8N{f_2} + T = 8{{N}}
By substituting 6N6{{N}} for f2{f_2} in the equation f2+T=8N{f_2} + T = 8{{N}}, we get,
6N+T=8N T=8N6N =2N 6{{N}} + T = 8{{N}}\\\ \Rightarrow {{T}} = 8{{N}} - 6{{N}}\\\ = 2{{N}}
Balance the force on 3kg3{{kg}} block to find the friction force f2{f_2}.
T+f2=1NT + {f_2} = 1{{N}}
By substituting 2N2{{N}} for TT in the equation T+f2=1NT + {f_2} = 1{{N}}, we get,
2N+f1=1N f1=1N2N =1N 2{{N}} + {f_1} = 1{{N}} \\\ \Rightarrow {f_1} = 1{{N}} - 2{{N}} \\\ = - 1{{N}}
The sign is negative so change the direction of f1{f_1} as shown below.

Therefore, the values of f1{f_1}, f2{f_2} and TT are 1N1{{N}}, 6N6{{N}}, and 2N2{{N}}, respectively.

Note: In this problem, the acceleration due to gravity is not given so take g=10m/s2g = 10{{m/}}{{{s}}^{{2}}}. The key part of this problem is a free body diagram which should be drawn in the correct manner which means the direction of each force will be correct otherwise the solution will be wrong.