## What is the required force to pull a shopping car?

A woman pulls a shopping car with a force F at an angle of 35. The car mass is 20kg and its friction coefficient is 0.05. Find:
(A) The force the woman needs to apply to move the car with a constant velocity of 1m/s.
(B) The magnitude of the normal force and friction when the car moves at constant velocity.
(C) The force the woman needs to exert to push the car with a constant velocity of 1/s.
(D) The Displacement if the car is pushed 5 seconds and the displacement if the car is pulled 5 seconds.

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### 3 Responses to “What is the required force to pull a shopping car?”

1. Itzel said :

The formula for friction is

Fr = μR

Where μ is the coefficient of friction, and
R is the normal contact force.

So the normal contact force will be
Mass x gravity = 50*9.81 = 490.5

0.37 * 490.5 = 181.485 N

It is worth noting that that answere is too precise and also that I ignored the vectors and just used magnitude, in more difficult problems you will need to use vectors.

2. Questor said :

……..Fy………F
…W…↑…….⁄
….↓….|….⁄_35°
._▓▓._|.⁄——→Fx
╘O=O╛←――Ff
▒▒↑▒▒▒▒▒▒▒▒▒▒▒▒
…..|
….N.

Given:
W = mg = 20kg(9.8)
µ = 0.05

(A) Force needed to PULL the car at a constant speed.
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
Summation of forces along the Y-axis = 0
∑Fy = 0
0 = -W + N + Fy
…= -mg + N + Fsin35°
N = mg – Fsin35° ………..◄eq1

Summation of forces along the X-axis = 0
∑Fx = 0
0 = Fx – Ff
Ff = Fx
Ff = Fcos35°

But Ff = µN …<==substitute value of N from ◄eq1 Fcos35° = µ [ mg - Fsin35°] F(.819) = 0.05[20(9.80) - F(.573)] .............= 9.8 - F(0.029) F(0.848) = 9.8 F = 11.56N (B) Normal force N = mg - Fsin35° ...........◄eq1 ...= 20 (9.80) - 11.56(.573) N = 196 - 6.622 N = 189.38N (C) WHEN PUSHING the shopping car ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ........Fy.........F ...W...|.......⁄ ....↓....|....⁄_35° ._▓▓._↓.⁄←------Fx ╘O=O╛――→Ff ▒▒↑▒▒▒▒▒▒▒▒▒▒▒▒ .....| ....N. Summation of forces along the Y-axis = 0 ∑Fy = 0 0 = -W + N - Fy ...= -mg + N Fsin35° N = mg + Fsin35° ...........◄eq1b Summation of forces along the X-axis = 0 ∑Fx = 0 0 = - Fx + Ff Ff = Fx Ff = Fcos35° But Ff = µN ...<==substitute value of N from ◄eq1b Fcos35° = µ [ mg + Fsin35°] F(.819) = 0.05[20(9.80) + F(.573)] .............= 9.8 + F(0.029) F(0.79) = 9.8 F = 12.40 N (D) Displacement whether being pulled or pushed will be the same since there is no Unbalanced Force because the velocity is constant at 1m/sec. Hence there is also no acceleration involved ......._ S = Vt S = (1m/sec)5 S = 5m (the same for pulling or pushing)

3. Matt said :

for the diagram—
http://rapidshare.com/files/295889546/Untitled.jpg.html
[only ten people can dwnld]

this is the hint
if u are able to solve it then exellent.

Message:

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