# Homework Help: Forces problem( help needed)

1. Feb 18, 2008

### Cathartics

forces problem(urgent urgent help needed)

HI there everyone, I am having trouble with this problems please help me with a good explanation and correct answer. Thanks in advance.

A 15 kg block of steel is at rest on a horizontal table. The coefficient of static friction between block and table is 0.50.
(a) What is the magnitude of the horizontal force that will put the block on the verge of moving?
answer in N

(b) What is the magnitude of a force acting upward 60° from the horizontal that will put the block on the verge of moving?
answer in N

(c) If the force acts down at 60° from the horizontal, how large can its magnitude be without causing the block to move?
answer in N

2. Feb 18, 2008

### Hootenanny

Staff Emeritus
Come on Carthics, you've been here long enough now to know that you need to show your attempts...

3. Feb 18, 2008

### Cathartics

THanks but no thanks hootenanny! I got the answer. And i am a beliver of live and let live and let others prosper.. Thats why i will post the solution here.. for other people who might need it.

a) Force of frcition,
f = μ.N = μmg = 0.5x15x9.8 = 73.5 N
(b) The vertical component of the force, upward reduces effective thrust down.
Normal reaction, N = mg - F sin θ
Force of limiting friction,
fs = μN = 0.5x15x9.8 - 0.5xF sin 60
= 73.5 - 0.433F
This must be equal to the horizontal component of the applied force,
F cos 60 = 0.5 F
Therefore,
0.5 F = 73.5 - 0.433 F
0.933 F = 73.5
F = 73.5/0.933 = 78.8 N

(c) The vertical component of the force, downward increases effective thrust down.
Normal reaction, N = mg + F sin θ
Force of limiting friction,
f's = μN' = 0.5x15x9.8 + 0.5xF sin 60
= 73.5 + 0.433F
This must be equal to the horizontal component of the applied force,
F cos 60 = 0.5 F
Therefore,
0.5 F = 73.5 + 0.433 F
0.067 F = 73.5
F = 73.5/0.067 = 1097 N

4. Feb 19, 2008

### Hootenanny

Staff Emeritus
Well done. However, you managed to work out the solution three hours after posting, so it begs that question, why did you really ask for help? Did you actually need help, or could you just not be bothered to attempt it yourself?

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