two physics questions, if you could help me step by step that'd be great since my teacher doesn't teach.

Question

1) A clerk moves a box of cans down an aisle by pulling on a rope attached to the box. The clerk pulls with a force of 188 N at an angle of 25.0° with the horizontal. The box has a mass of 36.9 kg, and the coefficient of kinetic friction between box and floor is 0.450. Find the acceleration of the box.

2)A box of books weighing 259 N is shoved across the floor by a force of 435 N exerted downward at an angle of 35° below the horizontal.

(a) If µk between the box and the floor is 0.57, how long does it take to move the box 9 m, starting from rest? (If the box will not move, enter 0.)

(b) If µk between the box and the floor is 0.75, how long does it take to move the box 7 m, starting from rest? (If the box will not move, enter 0.)

--what do the symbols even mean.... so yeah im really lost if u guys could explain how to do these probs so i know in the future, i'd appreciate it a lot. thanks

Answer 1

1) The angle of the rope needs to be considered to calculate horizontal and vertical components of the clerk's force.
Fh = 188*cos(25) = 170 N This is what pulls the box.
Fv = 188*sin(25) = 79.5 N This actually lifts the box slightly.

The symbol for coefficient of kinetic friction is µk. A lower number makes it easier to move the box - if you spilled some olive oil on the floor µk would perhaps go to 0.1. N is normal force. If this was on a slope, normal would be different from weight, but the floor here is level so N = W?

Actually no. Because the rope pulls up on the box slightly, the normal force pressing the box to the floor is less than the box's full weight. The box's weight = m*g = 36.9 kg*9.8 m/s^2 = 362 N. The vertical component of the tension in the rope relieves 79.5 N of that so N = 362 - 79.5 N = 282 N.

So Ff = .450*282 N = 127 N is the force that friction fights with. The horizontal portion of the clerk's pull is 170 N. So there is 43 N left over to accelerate the box. Newton's 2nd law says
F = m*a
43 N = 36.9 kg*a
so a = 1.17 m/s^2

2) Here the force is directed down at an angle so the vertical component of the force will add to the normal force used in calculating the friction.
Fh = 435 N*cos(35) = 356 N
Fv = 435 N*sin(35) = 250 N

N = the weight of the box + Fv = 259 N + 250 N = 509 N

a) Ff = µk*N = .57* 509 N = 290 N This is not enough to hold the box stationary so 356 - 290 N = 66 N are available to accelerate the box. Newton's 2nd law again:
But first we need the box's mass. W = m*g
m = W/g = 259/9.8 N = 26.4 kg. Now
F = m*a
66 N = 26.4 kg*a
a = 66/26.4 m/s^2 = 2.5 m/s^2

This is getting complicated. To move 9 m will take t seconds:
x = (1/2)*a*t^2 Solve for t.

b) Ff = µk*N = .75* 509 N = 381 N This is more than the horizontal portion of the clerk's push, so the box will resist. The answer is 0.

Answer 2

1) Given: Mass of box, m=36.9kg
If you multiply that by g, we will get its weight. Now g=9.8m/s^2, therefore weight of box,mg=36.9*9.8=361.6 N (That symbol means Newton, which is a unit of force. Weight is a form of force.)
Force applied on box at 25degrees=188N. This force has a vertical and horizontal component.
Vertical component=188sin25=
188*0.423=79.5N
Horizontal component=188cos25=
188*0.906=170.3N
Get a scientific calculator. This will give you the values of sin and cos of an angle which we showed above.

Coefficient of kinetic friction, uk=0.450

To find the acceleration of the box, we use the formula:

F=ma where F is the net force, m the mass of the box, and a the acceleration.

To get the net force, F, we get the value of the force of kinetic friction, f=ukR, where R is the total force normal to the floor. That R is equal to the weight of the box minus the vertical component of the applied force. Thus R=361.6-79.5=282.1N.

Therefore, f=0.450*282.1
=126.9N

The net force, F, is the difference between the force of friction,f, and the horizontal component of the applied force.

Thus F=170.3-126.9=43.4N.

Now we are ready to solve for the acceleration a. Substitute known values in our formula F=ma:

43.4=36.9a
a=43.4/36.9
=1.18m/s^2

2)In this problem, you need to memorize another formula:
s=ut+1/2at^2. Look up the meaning of each term in your textbook. The other formula which you need to enable you to solve this problem is the one we used in 1): F=ma. Solve for a as we did in 1). Take note that now the vertical component of the applied force is downward because the applied force is downward at an angle of 35degrees from the horizontal. Thus in computing for R you should add the vertical component of the applied force to the weight of the box.

The rest will be the same as in 1) above, except that you have to take another step by using s=ut+1/2at^2 to solve for t. Note that u=0 because the box starts from rest.

The other thing you have to see is: Is the force of friction,f greater than the horizontal component of the applied force? If yes, then the box will not move.

Answer 3

[1]
The force exerted is 188N * cos 25° = 170.4
The friction experienced by the box is µk*mg
=0.450 * 36.9 kg * 9.8 m/s^2
=162.729
Subtracting these forces,
170.4 - 162.73 = 7.66 N
Since the mass is 36.9 kg,
36.9 kg * a = 7.66 N
a = 0.20750308842520848481300540568595

[2]
(a) If the force exerted is same,
µk * m * g [9.8 m/s^2]
Frictional force = 206.1234
Therefore, the box will not move (0) .

(b)
Frictional force = 271.215
The friction is even greater than (a) !!
The box will not move (0).

µ = coefficient of friction (s = static; k = kinetic)
You multiply this constant to the weight of an object to find its frictional force.
g = acceleration due to gravity = 9.8 m/s/s
You multiply this constant to the mass of an object (kg) to find its weight (N).