A 120 kg crate is being moved from one location to another in a warehouse. The m

Question

A 120 kg crate is being moved from one location to another in a warehouse. The moment of inertia of a box rotating about its center of mass is (1/12) m [a^2 + b^2]. The crate is a square box, with each side having a length of 1.75 m. Forces are applied to the box as shown in the figure. What is the net force on the box? What is the net torque on the box? If the crate starts from rest, how long does it take to move the box 15.0 m? If the crate is allowed to rotate through an axis perpendicular to the ground along its center of mass, what is the angle that the crate will rotate through in this time?

Explanation / Answer

a)net force on the crate=200+300-230=270N

b) let us calculate torque due to all force with reference to the center of mass of the crate. as the crate is symmetric the center of mass lies at its geometric center ie at the intersection of the diagonals.

torque=forceX vertical component of the distance between the force and center of mass.

now net torque=200X1.75/2-300X1.75/2+230X0=-87.5Nm( - ve )sign ondicates that the crate will move in anticlockwise direction when it is allowed to rotate.

c) using the equation s=ut+1/2X(acc)X(time)2

  0+1/2(270/120)Xt2=15 solving time=3.6514sec

d) moment of inertiaX angle=net torque

so angle=(-87.5Nm)/massX(a2+b2/12)=(-87.5Nm)/61.25kgm2=-1.4285rad

( - ve )sign ondicates that the crate will move in anticlockwise direction when it is allowed to rotate.

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