The clerk pulls with a force of 194 N at an angle of 25.0° with the horizontal. The box has a mass of 35.4 kg, and the coefficient of kinetic friction between box and floor is 0.450. Find the acceleration of the box in m/s^2
2007-12-28 02:12:34 · 3 answers · asked by P 1 in Science & Mathematics ➔ Physics
The acceleration is in the horizontal direction. So, we need to find the forces acting in the this direction.
The forces are the tension in the rope and the frictional force.
The force of tension is straight forward to calculate. It is F cosθ. So, this turns out to be 194cos 25 = 175.82 N
The force of friction is μN, where N is the normal force. The normal force, the vertical component of the tension and the weight of the box balance each other; so the normal force = Weight-Fsin θ
= 35.4*9.8-194sin25 = 264.93 N
The force of friction is μ*264.93 = 119.21N
So, the net force in the horizontal direction is 175.82-119.21 = 56.61 N
The acceleration is therefore = F/m = 56.61/35.4 = 1.599 m/s2
2007-12-28 02:32:44 · answer #1 · answered by Ajinkya N 5 · 5⤊ 1⤋
1. The 194 N pull can be separated into two vectors:
A horizontal one 194*Cos(25) that will pull horizontally on the box (we'll need that later).
A vertical one 194*Sin(25) that will reduce the weight of the box upon the floor (we need that to calculate the friction)
2. The force of friction (trying to keep the box from moving) is the weight of the box (in Newtons) times the coefficient of friction.
The weight of the box is the gravity pulling on the box (35.4 kg * 9.8 m/s^2 = X Newtons) minus the upwards pull on the rope (found above).
The force of friction is this difference multiplied by the coefficient 0.450 -- this can be seen as the force with which the floor pulls back on the box (against the pull on the rope).
3. The acceleration will be due to the "net" horizontal force on the rope, after taking away what is needed to fight friction.
F = m a
a = F/m
F = horizontal pull on rope (found in 1) minus force of friction (found in 2).
m = 35.4 kg
Newtons divided by kg gives m/s^2
2007-12-28 02:27:33 · answer #2 · answered by Raymond 7 · 1⤊ 0⤋
Fnet = ma
Fa(cosθ) - Ff = ma
Fa(cosθ) - μkFn = ma
Fa(cosθ) - μk(mg-Fasinθ) = ma
a = [Fa(cosθ) - μk(mg-Fasinθ)]/m
a = [194(cos25.0° - 0.450{(35.4)(9.8)-194sin25.0°}]/35.4
a = 1.60 m/s^2 ANS
Hope I help you.
teddy boy
2007-12-28 02:41:30 · answer #3 · answered by teddy boy 6 · 1⤊ 0⤋