You are pushing on a door with a horizontal force of 300 N T

You are pushing on a door with a horizontal force of 300 N. The moment arm of this force around the hinges of the door is 45 cm (0.45 m). The other person is pushing in the opposite direction on the other side of the door (F_1). The moment arm of the other person\'s pushing force is 55 cm (0.55 m). How large is the force that the other person pushes with if the door is in static equilibrium? 366.7 N 245.45 N 300 N 24.54 N Which of the following energies make up the work - energy relationship? Gravitational potential energy Kinetic energy Strain energy Which of the following is a location in space? Position Velocity Speed Acceleration Displacement Which of the following forces result in the motion of the whole body? External Forces Internal Forces Compressive Forces Tensile Forces

Solution

QUESTION 30.

Horizontal force applied by first person(F1)=300 N

Moment arm of this force=0.45 m

Therefore, total magnitude of torque applied=(300*0.45)N = 135 N

Horizontal force applied by second person is F2 N.

Moment arm of this force=0.55 m

As the door is in static equilibrium,the net torque on the door is zero.

So,

135=F2*0.55

or, F2=245.45 N

Thus the other person pushes with a force of 245.45 N to keep the door in equilibrium.

QUESTION 31,32 and 33 are asked from a different subject and do not concern anything with algebra.