<span>We can draw a free body diagram for the safe to show all the horizontal forces acting on the safe. The three horizontal forces are Clyde's force, Bonnie's force, and friction. Since the safe slides at a constant speed (no acceleration), the net force must be zero.
0 = Clyde's force + Bonnie's force - friction
friction = Clyde's force + Bonnie's force
mg mu = Clyde's force + Bonnie's force
mu = (Clyde's force + Bonnie's force) / (mg)
mu = (445 N + 350 N) / (300 kg x 9.80~m/s^2)
mu = 0.27
The coefficient of friction is 0.27</span>
As in an element?
He/ Helium
<span>Weight of block, Wb = mass*gravity = 50*9.8 = 490 N</span>
Since block is being pulled up by a 13-degree slope
Therefore, Force which is acting parallel to the slop:
<span> F p =490 Sin </span>= 110.2N
Force which is acting perpendicular to the slope:
<span> Fv =490 Cos</span> = 477.4 N
Net force can be given as follows:
<span>F n = (250 - 110.2 - 0.2*</span>477.4) N
<span>Fn=44.3N</span>
Now acceleration is given by the ratio of force to mass
<span>a = Fn/m</span>
<span>=44.3/50 = 0.89 ms^<span>-2</span></span>
Answer:
Hey mate
Your answer will be <u>5 N</u><u> </u><u>Upwards</u><u>.</u><u> </u>
I will explain this to you.
In the attachment we can see that there are two forces acting on the object from the same side. And we know that when force acts from the same direction. Then the total force
=> sum of all the forces in that particular direction.
I hope you got it :)
TheSarcasmic