# M1 resolving forces question - help!Watch

#1
Question:

A box of mass 30kg is being pulled along rough horizontal ground at a constant speed using a rope. The rope makes an angle of 20° with the ground. The coefficient of friction between the box and the ground is 0.4. The box is modelled as a particle and the rope as a light, inextensible string. The tension in the rope is P newtons.

(a) Find the value of P

Okay, so here's my problem:

First, I drew a diagram to show the direction of all forces, including P, Friction, the resistant force amd 30g going down.

Here's my working out:

R(vertically) = 0
R + Psin20 = 30g (--This bit is the same as the mark scheme)
R(horizontally) = ma
Pcos20 - 0.4R(i.e. friction, with friction being μR) = 30a

However, the mark scheme has resolved horizontally differently; it has instead done Pcos20 = μR which I do not understand, because why hasn't the "ma" been taken into consideration?

The answer in the end should come out to be 110. It's an 8 mark question so probably needs a lot of working out?

If you could help me out with my dilemma, I'd be ever so grateful.
Thank you!
0
8 years ago
#2
Constant speed, so a=0
0
8 years ago
#3
(Original post by noobie93)
...
A constant speed means that acceleration is equal to 0.
0
#4
Ahh that would make sense! Thank you so much guys
0
8 years ago
#5
(Original post by noobie93)
Question:

A box of mass 30kg is being pulled along rough horizontal ground at a constant speed using a rope. The rope makes an angle of 20° with the ground. The coefficient of friction between the box and the ground is 0.4. The box is modelled as a particle and the rope as a light, inextensible string. The tension in the rope is P newtons.

(a) Find the value of P

Okay, so here's my problem:

First, I drew a diagram to show the direction of all forces, including P, Friction, the resistant force amd 30g going down.

Here's my working out:

R(vertically) = 0
R + Psin20 = 30g (--This bit is the same as the mark scheme)
R(horizontally) = ma
Pcos20 - 0.4R(i.e. friction, with friction being μR) = 30a

However, the mark scheme has resolved horizontally differently; it has instead done Pcos20 = μR which I do not understand, because why hasn't the "ma" been taken into consideration?

The answer in the end should come out to be 110. It's an 8 mark question so probably needs a lot of working out?

If you could help me out with my dilemma, I'd be ever so grateful.
Thank you!
There are 4 forces: friction, P, weight (mg), and normal reaction (you didn't mention it before)

horizontal Pcos20 - F = ma, constant speed so a = 0. F = μR
=> Pcos20 = μR (1)

Vertical R + Psin20 = W
=> R = mg - Psin20 (2)

sub (2) into (1)
Pcos20 = μ(mg - Psin20)
P(cos20 + μsin20) = μmg
P = μmg/(cos20 + μsin20) = 109N (3sf)

Hope this helps
0
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