total forces acting downward on the plane

Jun 2014
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the total force acting downwards the plane = 100(9.81)sin20 + 150= 485.5 N , should I take the tension in B into consideration ? how to get the tension on the rope ?
 

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ChipB

PHF Helper
Jun 2010
2,367
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Morristown, NJ USA
You can get the tension in the rope from the fact that point B is in equilibrium. Hence the sum of forces in the vertical direction at that point = 0.

Once you have the tension you can determine whether the sum of forces on the block acting in the direction of the incline is more than enough to cause it to slide, and if so - what the resulting sliding friction force is.
 
Jun 2014
306
0
You can get the tension in the rope from the fact that point B is in equilibrium. Hence the sum of forces in the vertical direction at that point = 0.

Once you have the tension you can determine whether the sum of forces on the block acting in the direction of the incline is more than enough to cause it to slide, and if so - what the resulting sliding friction force is.
if I take tension at B into consideration , i would have 100(9.81)sin20 +150+196.2 = 681.7N ??
 

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ChipB

PHF Helper
Jun 2010
2,367
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Morristown, NJ USA
No. Consider the sum of forces in y-direction at point B. You have 40 Kg x 9.81 m/s^2 acting downwards, so the tension in the rope must be:

T x sin(30) = 40 x 9.81,

T = 784.8 N
 
Last edited:
Jun 2014
306
0
No. Consider the sum of forces in y-direction at point B. You have 40 Kg x 9.81 m/s^2 acting downwards, so the tension in the rope must be:

T x sin(30) = 40 x 9.81,

T = 784.8 N
so, the total force = 100(9.81)sin20 +150+ 784.8 = 1270N ?
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
so, the total force = 100(9.81)sin20 +150+ 784.8 = 1270N ?
Two issues:

1. You left out the force of friction
2. You seem to have defined positive forces as pulling to the left (down the slope) - but which way is the tension on the rope acting on the block? Is it pushing to the left, or pulling to the right?
 
Jun 2014
306
0
Two issues:

1. You left out the force of friction
2. You seem to have defined positive forces as pulling to the left (down the slope) - but which way is the tension on the rope acting on the block? Is it pushing to the left, or pulling to the right?
1.) do u mean static force of friction ? it is 0.29x100x9.81xcos 20 = 276.5n

2.)i have found that the total force downards = 1270n as in previous post , can you correct my mistake ? i have no idea of what u 're talking about
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
You have defined positive direction as moving to the left. That's OK, but placing a plus sign in front of the value for tension in the rope means that you think is pushing to the left. But it can't - it's a rope! It must be in tension, trying to pull the block to the right - so the sign of the 784N force, due to tension in the rope, should negative.

I strongly suggest that for this problem, and the others you have posted recently, you should ALWAYS draw a free-body diagram of forces acting on the body of interest. This will help you see which way forces are acting on the body, and will allow you to determine the correct sign to use for each force.

Total force acting down the slope ignoring friction for the moment is

150 + 100*9.81*sin 20 - 784 = -299.3

The negative sign says that the sum of these forces is actually pulling to the right. Meaning the block will try to slide uphill, IF it can overcome static friction.

Next: is this force enough to overcome static friction?

F_s = 0.2 *100 * 9.81 *cos(20) = 276.6N. Since the upward force is greater than this, the block is moving. Now we can use sliding friction in the sum of forces:

Total F = -299.3 + 0.2 x 100 x 9.81 x cos(20) = -299.3 + 184.4 = -114.9N. Again, the minus sign says the sum of forces is upward, and the block slides up the ramp.
 
Last edited:
Jun 2014
306
0
You have defined positive direction as moving to the left. That's OK, but placing a plus sign in front of the value for tension in the rope means that you think is pushing to the left. But it can't - it's a rope! It must be in tension, trying to pull the block to the right - so the sign of the 784N force, due to tension in the rope, should negative.

I strongly suggest that for this problem, and the others you have posted recently, you should ALWAYS draw a free-body diagram of forces acting on the body of interest. This will help you see which way forces are acting on the body, and will allow you to determine the correct sign to use for each force.

Total force acting down the slope ignoring friction for the moment is

150 + 100*9.81*sin 20 - 784 = -299.3

The negative sign says that the sum of these forces is actually pulling to the right. Meaning the block will try to slide uphill, IF it can overcome static friction.

Next: is this force enough to overcome static friction?

F_s = 0.2 *100 * 9.81 *cos(20) = 276.6N. Since the upward force is greater than this, the block is moving. Now we can use sliding friction in the sum of forces:

Total F = -299.3 + 0.2 x 100 x 9.81 x cos(20) = -299.3 + 184.4 = -114.9N. Again, the minus sign says the sum of forces is upward, and the block slides up the ramp.
do u mean the force at right of the roller act to the left , so the reaction at the left of the roller should act to the right , in order for the roller to be in equilibrium?
 

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ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
The tension in the rope is equal on both sides of the roller, if that's what you are asking.