 # Mech 2 Chapter 7 - Applications of Forces

Designed to accompany the Pearson Applied Mathematics Year 2/AS textbook. ## Mr A Blackett

### 16th Jan 2020 Flag Comment

slide 18 resolving down the plane should be 3/5 mg, not 3/4 mg. Thanks as every for your exceptional work. ## Dr J Frost

### 7th Jan 2019 Flag Comment

I've resolved this problem by changing the force. ## Mr z huang

### 19th Dec 2018 Flag Comment

Direction of force Q should NOT point as the diagram shown. since to balance out 4N Up and p 6N* cos(30) Down the plane,(both perpendicular). WE DO NEED to force points UP. otherwise the resultant force on the direction perpendicular to the plane cannot be 0 ## Mr z huang

### 19th Dec 2018 Flag Comment

Direction of force Q should NOT point as the diagram shown. since to balance out 4N Up and p 6N* cos(30) Down the plane,(both perpendicular). WE DO NEED to force points UP. otherwise the resultant force on the direction perpendicular to the plane cannot be 0. ## Mr G Richmond

### 20th Nov 2018 Flag Comment

Can anyone please help. On slide 6, the solution to the angle is 21.4 degrees. I assume this is an error as tan 21.4 does not equal the division of (4-6cos30 / 6sin30 + 5). However, I am not sure of what the answer should be. When I solve it I get an answer of -8.50 degrees. This doesn't make sense to me either, why would I get a negative answer? If I use tan graphs and add 180 that gives me 171.5 degrees, which it also cant be given that its effectively an angle in a right angled triangle. Any ideas?????? Thanks.

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