You pull with a force of 235 N on a rope that is attached to a block of mass 28 kg, and the block slides across the floor at a constant speed of 1.7 m/s.

Question

Q: You pull with a force of 235 N on a rope that is attached to a block of mass 28 kg, and the block slides across the floor at a constant speed of 1.7 m/s. The rope makes an angle of α= 46 degrees with the horizontal

a) What is the net force on the block?

b) Which objects exert forces on the block with nonzero x-components?

options: floor, earth, you, rope. I was thinking it was floor and earth

c) What is the x-component FT, x of the tension force exerted by the rope on the block? (A component may be positive or negative.)

d) What is the x-component Ff,x of the force exerted by the floor on the block (the friction force)?

e) Which objects exert forces on the block with nonzero y-components?

options: floor, earth, you, rope

f) What is the y-component FT, y of the force exerted by the rope on the block?

g) What is the y-component FEarth, y of the force exerted by the Earth on the block?

h) What is the y-component FN, y of the force exerted by the floor on the block (sometimes called the "normal" force, because it is perpendicular to the floor)?

Answer 1

I assume that the direction of movement is considered positive x-direction, and

that the upward is the positive y-direction.

a)

The block moves with constant speed,

therefore its acceleration is 0,

therefore the net force is 0.

b) and e)

The floor exerts two forces:

i) The normal force -- reaction to the block's weight.

The normal force is vertically up, so

its x component is 0, and y component is non-0.

ii) The force of friction, which is opposite to movement, so

its x component is non-0, and y component is 0.

The Earth exerts the force of gravity, which is vertically down, so

its x component is 0, and y component is non-0.

The rope exerts a force at an angle of 46°, so

both x and y component are non-0.

Concerning the force exerted by you on the block, it depends on the definition

of "exert". The thing is that you exert force on the rope and the rope exerts

a force on the block.

By my definition of "exert" you do not exert any force on the block.

But I am not sure what definition the teacher gave you; it is possible that

because you exert a force via the string he/she expect the same answer as for the rope.

c)

On the assumption that the string has no mass, the string's tension is the same as

the force you apply -- 235 N.

And the x-component is the cosine of the angle, that is,

235cos(46°)

d)

The net force is 0, and the force of friction is the only one that can cancel out

the x-component of the string's tension.

Therefore the force of friction is -235cos(46°).

f)

On the assumption that the string has no mass, the string's tension is the same as

the force you apply -- 235 N.

And the y-component is the sine of the angle, that is,

235sin(46°)

g)

The earth exerts the downward (i.e. negative) force of gravity of magnitude

-28g

where g is the gravitational acceleration, approximately 9.81 m/s².

h)

The normal force counteracts the block's weight (question g), with the "help" of

the rope's tension (question f).

Therefore the normal force is

28g - 235sin(46°)