1. ## forces

a block of mass 20kg is held in equilibrium on a plane by means of a string which is at an an angle of 25° to the greatest slope of the plane. the plane is at 40° to the horizontal.
show that the tension in the string is about 142N
i thought this would be pretty straightforward but i just don't get 142N! i mostly used 200cos60° = Tcos25° but this doesn't wrok, why? are there forces i am missing? no friction btw

2. Originally Posted by furor celtica
a block of mass 20kg is held in equilibrium on a plane by means of a string which is at an an angle of 25° to the greatest slope of the plane. the plane is at 40° to the horizontal.
show that the tension in the string is about 142N
i thought this would be pretty straightforward but i just don't get 142N! i mostly used 200cos60° = Tcos25° but this doesn't wrok, why? are there forces i am missing? no friction btw
The forces acting on the block are (1) its weight (vertically downwards), (2) the tension T in the string (at an angle 25º to the plane), (3) the normal force N supporting the block on the plane (acting perpendicular to the plane because we're assuming there's no friction).

To calculate T, resolve the forces in the direction of the slope of the plane. Then the effect of N is eliminated, the weight makes an angle 50º (not 60º!) to this direction, and the tension makes an angle 25º. So the equation is $200\cos50^\circ = T\cos25^\circ$.