Question 13 Marks
A person (40kg) is managing to be at rest between two vertical walls by pressing one wall A by his hands and feet and the other wall B by his back. Assume that the friction coefficient between his body and the walls is 0.8 and that limiting friction acts at all the contacts.
- Show that the person pushes the two walls with equal force.
- Find the normal force exerted by either wall on the person. Take $g = 10m/s^2.$
Answer
Frictional force developed between hands, legs & back side with the wall the wt of man. So he remains in equilibrium.
He gives equal force on both the walls so gets equal reaction R from both the walls. If he applies unequal forces R should be different he can’t rest between the walls.
Frictional force $2\mu\text{R}$ balance his wt.
From the free body diagram
$\mu\text{R}+\mu\text{R = 40g}$
$\Rightarrow2\mu\text{R}=40\times10\Rightarrow\text{R}=\frac{40\times10}{2\times0.8}=250\text{N}$
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- Mass of man = 50kg.
Frictional force developed between hands, legs & back side with the wall the wt of man. So he remains in equilibrium.
He gives equal force on both the walls so gets equal reaction R from both the walls. If he applies unequal forces R should be different he can’t rest between the walls.
Frictional force $2\mu\text{R}$ balance his wt.
From the free body diagram
$\mu\text{R}+\mu\text{R = 40g}$
$\Rightarrow2\mu\text{R}=40\times10\Rightarrow\text{R}=\frac{40\times10}{2\times0.8}=250\text{N}$
- The normal force is 250N.
Due to friction the body will decelerate