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M1 Resolving Forces

Can anyone help me with this question?

A sack of mass 8kg is being pulled up a ramp inclined 20 degrees to the horizontal by a force of magnitude 45N acting parallel to a line of greatest slope of the ramp. The acceleration of the sack is 1.4m/s^2. Find the frictional force.

Reply 1

HurtfulHarold

The first step to questions like these is to draw a good diagram with all of the relevant information on it. Could you perhaps provide us with a picture of your diagram?

Reply 2

CodgerCollins

Here's my interpretation. Not sure I've drawn in the Action force correctly though. A983DA79-2A9D-45A7-B3E6-D0191437CA61.jpg.jpeg

What does it mean by parallel to the line of greatest slope of the ramp?

34C8EA01-1153-4308-98F0-14726A2D9B1C.jpg.jpeg17.0KB

Reply 3

CodgerCollins

Just did some working out and came up with this
4B3BF32E-E27A-4E88-AC84-F72B7A188157.jpg.jpeg

Does this look right?

Reply 4

Muttley79

Original post by CodgerCollins

Just did some working out and came up with this

Does this look right?

You should be using F=ma parallel to the slope and taking up the slope as positive.

Forces are not equal as the sack is not in equilibrium.

Reply 5

CodgerCollins

Original post by Muttley79

You should be using F=ma parallel to the slope and taking up the slope as positive.

Forces are not equal as the sack is not in equilibrium.

Could you expand on what you mean by "F=ma parallel to the slope"?

Reply 6

Muttley79

Original post by CodgerCollins

Could you expand on what you mean by "F=ma parallel to the slope"?

Which force is pulling the sack up the slope? 45N

Which forces are resisting the motion? ..

Have a go and I'll give you the next hint.

Reply 7

CodgerCollins

So the resisting forces would be the frictional force, the weight(w(sin(20)). Would the normal contact force act as a resistive force??

Reply 8

Muttley79

Original post by CodgerCollins

So the resisting forces would be the frictional force, the weight(w(sin(20)). Would the normal contact force act as a resistive force??

Normal contact force is acting perpendicular to the slope so has no effect parallel to the slope.

Now we know the accel and the forces can you use F=ma now?