Physics Question(Simple but Explain in VERY GOOD DETAIL)?

Question

A man pulls a 50kg box at constant speed across the floor. He applies a 200N force at an angel of 30 degrees.

A. Sum the forces in the x-direction. What is the value of the frictional force opposing the motion?

B. Sum the forces in the y-direction. What is the value of the normal force?

Alright, it's not a horizontal plane. It gave a picture so I am going to try to copy it using character's.

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Answer 1

A. There are two forces acting on the box in the x-direction.
1. The frictional force exerted by the ground in a direction opposite to motion.
2. The horizontal component of the 200 N force he applies at an angle of 30 degrees from the horizontal.

Answer-1)Since the box moves at constant speed, both the above forces add up to a resultant of 0.
Answer-2) Value of frictional force opposing motion= 200* cos 30
= 173.2 N

B. There are three forces acting in the y- direction:
1. Normal Reaction exerted by the ground on the box.
2. Force of gravity on the box
3. Vertical component of the 200 N force.

Answer-3)Since the body does not shown any sign of acceleration in the vertical direction, the sum of all the above forces is also 0.
Answer-4) Normal Reaction(upwards) = 50 * 9.8 - 200* sin30
= 490 - 100 = 390N

The above holds if the angle of 30 degrees is wrt to the horizontal. Else if angle with horizontal is 60 degrees

Answer-1) 0
Answer-2) 200* Cos 60 = 100 N
Answer-3) 0
Answer-4) 50* 9.8 - 200*sin60 = 490 - 173.2 = 316.8 N

Answer 2

To find the friction force, it is easy to apply the rules parallel and perpendicular to the plane. When it is done, it can be seen that the friction force is 450 N.

As this man pulls the box at a constant speed, the sum of the forces in x-direction and sum of the forces in y-direction are zero (0). This is because, F=ma (F-force in a particular direction m-weight of the object a-acceleration of the object). In this case, F=0 because it is travelling at a constant speed.

The normal force is equals to 500*Cos(30deg)=433.0127 N

Answer 3

Hmm.. 30 degrees relative to what??

If it's relative to horizontal plane, then:
A. If the box move at constant speed (a=0) then the sigma Fx is zero.
sigma Fx=m.a
F-f=0
f=F
The force applied is 200N at an angle of 30 deg from horizontal plane, then F=200.cos30deg=100.square root(3)..
Hence, the friction will be roughly around 173.20 N.
B. The box is not being lifted, therefore the sigma Fy is equals to zero.
The force applied at vertical position is 200.sin30deg=100N.
The weight of the box: w=m.g=50.10=500N.
sigma Fy=m.a
F+N-w=0
100+N-500=0
N=400N
The normal force is 400N.


hope it helps
peace
vixklen

Answer 4

A. 56

B. 56

Answer 5

200N*cos(30deg)


50kg*g - 200N*sin(30deg)

Answer 6

You're right, it was pretty simple.

Answer 7

Yeap u got it...