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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Friction

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Friction

A person ( 40 kg ) manages to remain at rest between two vertical...

A person (40 kg40 \text{ kg}) manages to remain at rest between two vertical walls by pressing one wall AA with his hands and feet and the other wall BB with his back, as shown in the figure. The friction coefficient between his body and the walls is 0.80.8, and limiting friction acts at all contact points. Let FAF_A and FBF_B represent the forces with which the person pushes wall AA and wall BB, respectively. Which of the following relationships is correct?
A person ( 40 kg ) manages to remain at rest between two vertical walls by pressing one wall A with his hands and feet

Options

  1. AFA=2FBF_A = 2F_B
  2. BFB=2FAF_B = 2F_A
  3. CFA=FBF_A = F_B
  4. DFA=0.8FBF_A = 0.8 F_B

Correct answer

C. FA=FBF_A = F_B

Step-by-step solution

For the person to remain at rest, the net force acting on him in any direction must be zero.

Considering the horizontal direction, the only forces acting on the person are the normal reactions from the vertical walls:
1. Normal reaction NAN_A from wall A acting towards the left.
2. Normal reaction NBN_B from wall B acting towards the right.

For horizontal equilibrium, we must have:
∑Fx=0\sum F_x = 0
NB−NA=0N_B - N_A = 0
NA=NBN_A = N_B

By Newton's third law, the normal force the person exerts on wall A is FA=NAF_A = N_A, and the normal force he exerts on wall B is FB=NBF_B = N_B.

Therefore, it directly follows that:
FA=FBF_A = F_B
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